LIMITS AND CONTINUITY

Transcription

1 Chapter LIMITS AND CONTINUITY OVERVIEW The concept of a it is a central idea that distinguishes calculus from algebra and trigonometr. It is fundamental to finding the tangent to a curve or the velocit of an object. In this chapter we develop the it, first intuitivel and then formall. We use its to describe the wa a function ƒ varies. Some functions var continuousl; small changes in produce onl small changes in ƒ(). Other functions can have values that jump or var erraticall. The notion of it gives a precise wa to distinguish between these behaviors. The geometric application of using its to define the tangent to a curve leads at once to the important concept of the derivative of a function. The derivative, which we investigate thoroughl in Chapter 3, quantifies the wa a function s values change.. Rates of Change and Limits In this section, we introduce average and instantaneous rates of change. These lead to the main idea of the section, the idea of it. Average and Instantaneous Speed A moving bod s average speed during an interval of time is found b dividing the distance covered b the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet per second, or whatever is appropriate to the problem at hand. EXAMPLE Finding an Average Speed A rock breaks loose from the top of a tall cliff. What is its average speed (a) during the first sec of fall? (b) during the -sec interval between second and second? Solution In solving this problem we use the fact, discovered b Galileo in the late siteenth centur, that a solid object dropped from rest (not moving) to fall freel near the surface of the earth will fall a distance proportional to the square of the time it has been falling. (This assumes negligible air resistance to slow the object down and that gravit is 73

2 74 Chapter : Limits and Continuit HISTORICAL BIOGRAPHY* Galileo Galilei (564 64) the onl force acting on the falling bod. We call this tpe of motion free fall.) If denotes the distance fallen in feet after t seconds, then Galileo s law is where 6 is the constant of proportionalit. The average speed of the rock during a given time interval is the change in distance,, divided b the length of the time interval, t. (a) For the first sec: (b) From sec to sec : t t = 6t, = 6sd - 6sd - = 6sd - 6sd - = 3 ft sec = 48 ft sec The net eample eamines what happens when we look at the average speed of a falling object over shorter and shorter time intervals. EXAMPLE Finding an Instantaneous Speed Find the speed of the falling rock at t = and t = sec. Solution We can calculate the average speed of the rock over a time interval [t, t + h], having length t = h, as t = 6st + hd - 6t. h We cannot use this formula to calculate the instantaneous speed at t b substituting h =, because we cannot divide b zero. But we can use it to calculate average speeds over increasingl short time intervals starting at t = and t =. When we do so, we see a pattern (Table.). TABLE. Average speeds over short time intervals Average speed: t = 6st + hd - 6t h Length of Average speed over Average speed over time interval interval of length h interval of length h h starting at t starting at t () The average speed on intervals starting at t = seems to approach a iting value of 3 as the length of the interval decreases. This suggests that the rock is falling at a speed of 3 ft> sec at t = sec. Let s confirm this algebraicall. To learn more about the historical figures and the development of the major elements and topics of calculus, visit

3 . Rates of Change and Limits 75 If we set t = and then epand the numerator in Equation () and simplif, we find that t = 6s + hd - 6sd h = 3h + 6h h = 3 + 6h. For values of h different from, the epressions on the right and left are equivalent and the average speed is 3 + 6h ft>sec. We can now see wh the average speed has the iting value 3 + 6sd = 3 ft>sec as h approaches. Similarl, setting t = in Equation (), the procedure ields t = 6s + h + h d - 6 h = h for values of h different from. As h gets closer and closer to, the average speed at t = sec has the iting value 64 ft> sec. Average Rates of Change and Secant Lines Given an arbitrar function = ƒsd, we calculate the average rate of change of with respect to over the interval [, ] b dividing the change in the value of, = ƒs d - ƒs d, b the length = - = h of the interval over which the change occurs. DEFINITION Average Rate of Change over an Interval The average rate of change of = ƒsd with respect to over the interval [, ] is = ƒs d - ƒs d - = ƒs + hd - ƒs d, h Z. h f() Q(, f( )) Geometricall, the rate of change of ƒ over [, ] is the slope of the line through the points Ps, ƒs dd and Qs, ƒs dd (Figure.). In geometr, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of ƒ from to is identical with the slope of secant PQ. Eperimental biologists often want to know the rates at which populations grow under controlled laborator conditions. P(, f( )) Secant h FIGURE. A secant to the graph = ƒsd. Its slope is >, the average rate of change of ƒ over the interval [, ]. EXAMPLE 3 The Average Growth Rate of a Laborator Population Figure. shows how a population of fruit flies (Drosophila) grew in a 5-da eperiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time, and the points joined b a smooth curve (colored blue in Figure.). Find the average growth rate from da 3 to da 45. Solution There were 5 flies on da 3 and 34 flies on da 45. Thus the number of flies increased b 34-5 = 9 in 45-3 = das. The average rate of change of the population from da 3 to da 45 was Average rate of change: p t = = 9 L 8.6 flies>da.

4 76 Chapter : Limits and Continuit p Number of flies P(3, 5) p 8.6 flies/da t t Q(45, 34) p Time (das) t FIGURE. Growth of a fruit fl population in a controlled eperiment. The average rate of change over das is the slope p> t of the secant line. This average is the slope of the secant through the points P and Q on the graph in Figure.. The average rate of change from da 3 to da 45 calculated in Eample 3 does not tell us how fast the population was changing on da 3 itself. For that we need to eamine time intervals closer to the da in question. EXAMPLE 4 The Growth Rate on Da 3 How fast was the number of flies in the population of Eample 3 growing on da 3? Solution To answer this question, we eamine the average rates of change over increasingl short time intervals starting at da 3. In geometric terms, we find these rates b calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure.3). p Q (45, 34) (4, 33) (35, 3) (3, 65) Slope of PQ p/ t (flies/ da) L 8.6 L.6 L 3.3 L 6.4 Number of flies 35 B(35, 35) Q(45, 34) P(3, 5) A(4, ) Time (das) t FIGURE.3 The positions and slopes of four secants through the point P on the fruit fl graph (Eample 4).

5 . Rates of Change and Limits 77 The values in the table show that the secant slopes rise from 8.6 to 6.4 as the t-coordinate of Q decreases from 45 to 3, and we would epect the slopes to rise slightl higher as t continued on toward 3. Geometricall, the secants rotate about P and seem to approach the red line in the figure, a line that goes through P in the same direction that the curve goes through P. We will see that this line is called the tangent to the curve at P. Since the line appears to pass through the points (4, ) and (35, 35), it has slope = 6.7 flies>da (approimatel). On da 3 the population was increasing at a rate of about 6.7 flies> da. The rates at which the rock in Eample was falling at the instants t = and t = and the rate at which the population in Eample 4 was changing on da t = 3 are called instantaneous rates of change. As the eamples suggest, we find instantaneous rates as iting values of average rates. In Eample 4, we also pictured the tangent line to the population curve on da 3 as a iting position of secant lines. Instantaneous rates and tangent lines, intimatel connected, appear in man other contets. To talk about the two constructivel, and to understand the connection further, we need to investigate the process b which we determine iting values, or its, as we will soon call them. Limits of Function Values Our eamples have suggested the it idea. Let s begin with an informal definition of it, postponing the precise definition until we ve gained more insight. Let ƒ() be defined on an open interval about, ecept possibl at itself. If ƒ() gets arbitraril close to L (as close to L as we like) for all sufficientl close to, we sa that ƒ approaches the it L as approaches, and we write which is read the it of ƒ() as approaches is L. Essentiall, the definition sas that the values of ƒ() are close to the number L whenever is close to (on either side of ). This definition is informal because phrases like arbitraril close and sufficientl close are imprecise; their meaning depends on the contet. To a machinist manufacturing a piston, close ma mean within a few thousandths of an inch. To an astronomer studing distant galaies, close ma mean within a few thousand light-ears. The definition is clear enough, however, to enable us to recognize and evaluate its of specific functions. We will need the precise definition of Section.3, however, when we set out to prove theorems about its. EXAMPLE 5 How does the function behave near =? Behavior of a Function Near a Point Solution The given formula defines ƒ for all real numbers ecept = (we cannot divide b zero). For an Z, we can simplif the formula b factoring the numerator and canceling common factors: ƒsd = s - ds + d - ƒsd = L, : ƒsd = - - = + for Z.

6 78 Chapter : Limits and Continuit The graph of ƒ is thus the line = + with the point (, ) removed. This removed point is shown as a hole in Figure.4. Even though ƒ() is not defined, it is clear that we can make the value of ƒ() as close as we want to b choosing close enough to (Table.). f() TABLE. The closer gets to, the closer ƒ() ( )/( ) seems to get to Values of below and above ƒ(), FIGURE.4 The graph of ƒ is identical with the line = + ecept at =, where ƒ is not defined (Eample 5). We sa that ƒ() approaches the it as approaches, and write ƒsd =, or - : : - =. EXAMPLE 6 The Limit Value Does Not Depend on How the Function Is Defined at The function ƒ in Figure.5 has it as : even though ƒ is not defined at =. The function g has it as : even though Z gsd. The function h is the onl one (a) f(), (b) g(), (c) h() FIGURE.5 The its of ƒ(), g(), and h() all equal as approaches. However, onl h() has the same function value as its it at = (Eample 6).

7 . Rates of Change and Limits 79 whose it as : equals its value at =. For h, we have : hsd = hsd. This equalit of it and function value is special, and we return to it in Section.6. Sometimes : ƒsd can be evaluated b calculating ƒs d. This holds, for eample, whenever ƒ() is an algebraic combination of polnomials and trigonometric functions for which ƒs d is defined. (We will sa more about this in Sections. and.6.) EXAMPLE 7 Finding Limits b Calculating ƒ( ) (a) Identit function k k (a) (b) (c) (d) (e) s4d = 4 : s4d = 4 : -3 = 3 :3 s5-3d = - 3 = 7 : : = =- 3 (b) Constant function FIGURE.6 The functions in Eample 8. EXAMPLE 8 The Identit and Constant Functions Have Limits at Ever Point (a) If ƒ is the identit function ƒsd =, then for an value of (Figure.6a), (b) If ƒ is the constant function ƒsd = k (function with the constant value k), then for an value of (Figure.6b), For instance, ƒsd = =. : : ƒsd = k = k. : : = 3 and s4d = s4d = 4. :3 : -7 : We prove these results in Eample 3 in Section.3. Some was that its can fail to eist are illustrated in Figure.7 and described in the net eample.,,,,, sin, (a) Unit step function U() (b) g() (c) f() FIGURE.7 None of these functions has a it as approaches (Eample 9).

8 8 Chapter : Limits and Continuit EXAMPLE 9 A Function Ma Fail to Have a Limit at a Point in Its Domain Discuss the behavior of the following functions as :. (a) Usd = e, 6, Ú, Z (b) (c) gsd = L, = ƒsd =, sin, 7 Solution (a) It jumps: The unit step function U() has no it as : because its values jump at =. For negative values of arbitraril close to zero, Usd =. For positive values of arbitraril close to zero, Usd =. There is no single value L approached b U() as : (Figure.7a). (b) It grows too large to have a it: g() has no it as : because the values of g grow arbitraril large in absolute value as : and do not sta close to an real number (Figure.7b). (c) It oscillates too much to have a it: ƒ() has no it as : because the function s values oscillate between + and - in ever open interval containing. The values do not sta close to an one number as : (Figure.7c). Using Calculators and Computers to Estimate Limits Tables. and. illustrate using a calculator or computer to guess a it numericall as gets closer and closer to. That procedure would also be successful for the its of functions like those in Eample 7 (these are continuous functions and we stud them in Section.6). However, calculators and computers can give false values and misleading impressions for functions that are undefined at a point or fail to have a it there. The differential calculus will help us know when a calculator or computer is providing strange or ambiguous information about a function s behavior near some point (see Sections 4.4 and 4.6). For now, we simpl need to be attentive to the fact that pitfalls ma occur when using computing devices to guess the value of a it. Here s one eample. EXAMPLE Guess the value of Guessing a Limit + - :. Solution Table.3 lists values of the function for several values near =. As approaches through the values ;, ;.5, ;., and ;., the function seems to approach the number.5. As we take even smaller values of, ;.5, ;., ;., and ;., the function appears to approach the value. So what is the answer? Is it.5 or, or some other value? The calculator/computer values are ambiguous, but the theorems on its presented in the net section will confirm the correct it value to be.5a = B. Problems such as these demonstrate the

9 . Rates of Change and Limits 8 TABLE.3 Computer values of ƒ() + - Near ƒ() ; ; t approaches.5? ; ;..5 ;.5.8 ;.. t approaches? ;.. ;.. power of mathematical reasoning, once it is developed, over the conclusions we might draw from making a few observations. Both approaches have advantages and disadvantages in revealing nature s realities.

10 . Rates of Change and Limits 8 EXERCISES. Limits from Graphs. For the function g() graphed here, find the following its or eplain wh the do not eist. a. gsd b. gsd c. : : g() 3 gsd :3 3. Which of the following statements about the function graphed here are true, and which are false? a. eists. b. c. d. e. ƒsd : ƒsd =. : : : : ƒsd =. ƒsd =. ƒsd =. f. ƒsd eists at ever point in s -, d. : = ƒsd. For the function ƒ(t) graphed here, find the following its or eplain wh the do not eist. a. b. c. t: - ƒstd t: - ƒstd s f(t) s ƒstd t: t f() 4. Which of the following statements about the function graphed here are true, and which are false? a. ƒsd does not eist. : b. ƒsd =. : = ƒsd

11 8 Chapter : Limits and Continuit T ƒ ƒ ƒ ƒ ƒ ƒ c. does not eist. : 3. Let Gsd = s + 6d>s d. d. ƒsd eists at ever point in s -, d. a. Make a table of the values of G at = -5.9, -5.99, , : and so on. Then estimate :-6 Gsd. What estimate do ou e. ƒsd eists at ever point in (, 3). : arrive at if ou evaluate G at = -6., -6., -6., Á instead? f() b. Support our conclusions in part (a) b graphing G and using Zoom and Trace to estimate -values on the graph as : c. Find :-6 Gsd algebraicall. 4. Let hsd = s - - 3d>s d. a. Make a table of the values of h at =.9,.99,.999, and so on. Then estimate :3 hsd. What estimate do ou arrive at if ou evaluate h at = 3., 3., 3., Á instead? Eistence of Limits b. Support our conclusions in part (a) b graphing h near In Eercises 5 and 6, eplain wh the its do not eist. = 3 and using Zoom and Trace to estimate -values on the 5. graph as : : : - c. Find :3 hsd algebraicall. 7. Suppose that a function ƒ() is defined for all real values of ecept =. Can anthing be said about the eistence of 5. Let ƒsd = s - d>s - d. a. Make tables of the values of ƒ at values of that approach : ƒsd? Give reasons for our answer. = - from above and below. Then estimate :- ƒsd. 8. Suppose that a function ƒ() is defined for all in [-, ]. Can b. Support our conclusion in part (a) b graphing ƒ near anthing be said about the eistence of : ƒsd? Give reasons for our answer. = - and using Zoom and Trace to estimate -values on the graph as : If : ƒsd = 5, must ƒ be defined at =? If it is, must c. Find ƒsd = 5? Can we conclude anthing about the values of ƒ at :- ƒsd algebraicall. =? Eplain. 6. Let Fsd = s d>s - d.. If ƒsd = 5, must : ƒsd eist? If it does, then must a. Make tables of values of F at values of that approach : ƒsd = 5? Can we conclude anthing about : ƒsd? Eplain. = - from above and below. Then estimate :- Fsd. b. Support our conclusion in part (a) b graphing F near Estimating Limits = - and using Zoom and Trace to estimate -values on the graph as : -. You will find a graphing calculator useful for Eercises. c. Find. Let ƒsd = s :- Fsd algebraicall. - 9d>s + 3d. 7. Let gsud = ssin ud>u. a. Make a table of the values of ƒ at the points = -3., -3., -3., and so on as far as our calculator can go. a. Make a table of the values of g at values of u that approach Then estimate :-3 ƒsd. What estimate do ou arrive at if u = from above and below. Then estimate u: gsud. ou evaluate ƒ at = -.9, -.99, -.999, Á instead? b. Support our conclusion in part (a) b graphing g near b. Support our conclusions in part (a) b graphing ƒ near = -3 and using Zoom and Trace to estimate -values on the graph as : -3. c. Find :-3 ƒsd algebraicall, as in Eample 5.. Let gsd = s - d> A - B. a. Make a table of the values of g at the points =.4,.4,.44, and so on through successive decimal approimations of. Estimate : gsd. b. Support our conclusion in part (a) b graphing g near = and using Zoom and Trace to estimate -values on the graph as :. c. Find : gsd algebraicall. u =. 8. Let Gstd = s - cos td>t. a. Make tables of values of G at values of t that approach t = from above and below. Then estimate t: Gstd. b. Support our conclusion in part (a) b graphing G near t =. 9. Let ƒsd = >s - d. a. Make tables of values of ƒ at values of that approach = from above and below. Does ƒ appear to have a it as :? If so, what is it? If not, wh not? b. Support our conclusions in part (a) b graphing ƒ near =.

12 . Rates of Change and Limits 83. Let ƒsd = s3 - d>. a. Estimate the slopes of secants PQ, PQ, PQ 3, and PQ 4, a. Make tables of values of ƒ at values of that approach = arranging them in order in a table like the one in Figure.3. from above and below. Does ƒ appear to have a it as What are the appropriate units for these slopes? :? If so, what is it? If not, wh not? b. Support our conclusions in part (a) b graphing ƒ near b. Then estimate the Cobra s speed at time t = sec. 36. The accompaning figure shows the plot of distance fallen versus =. time for an object that fell from the lunar landing module a distance 8 m to the surface of the moon. Limits b Substitution In Eercises 8, find the its b substitution. Support our answers with a computer or calculator if available. a. Estimate the slopes of the secants PQ, PQ, PQ 3, and PQ 4, arranging them in a table like the one in Figure.3. b. About how fast was the object going when it hit the surface?. :. : - 3. s3 - d 4. 8 P :>3 : s3 - d Q s - d 6. 6 Q 3 : - : Q cos :p> :p - p Q Average Rates of Change In Eercises 9 34, find the average rate of change of the function over the given interval or intervals. 9. ƒsd = 3 + ; a. [, 3] b. [-, ] 3. gsd = ; a. [-, ] b. [-, ] 3. hstd = cot t; a. [p>4, 3p>4] b. [p>6, p>] 3. gstd = + cos t; a. [, p] b. [-p, p] Rsud = 4u + ; [, ] Psud = u 3-4u + 5u; [, ] 35. A Ford Mustang Cobra s speed The accompaning figure shows the time-to-distance graph for a 994 Ford Mustang Cobra accelerating from a standstill. Distance (m) s 5 Q Q Q 3 Q 4 5 Elapsed time (sec) P t T T T Distance fallen (m) 5 Elapsed time (sec) 37. The profits of a small compan for each of the first five ears of its operation are given in the following table: Year Profit in $s a. Plot points representing the profit as a function of ear, and join them b as smooth a curve as ou can. b. What is the average rate of increase of the profits between 99 and 994? c. Use our graph to estimate the rate at which the profits were changing in Make a table of values for the function Fsd = s + d>s - d at the points =., = >, = >, = >, = >, and =. a. Find the average rate of change of F() over the intervals [, ] for each Z in our table. b. Etending the table if necessar, tr to determine the rate of change of F() at =. 39. Let gsd = for Ú. a. Find the average rate of change of g() with respect to over the intervals [, ], [,.5] and [, + h]. b. Make a table of values of the average rate of change of g with respect to over the interval [, + h] for some values of h t

13 84 Chapter : Limits and Continuit T approaching zero, sa h =.,.,.,.,., and.. c. What does our table indicate is the rate of change of g() with respect to at =? d. Calculate the it as h approaches zero of the average rate of change of g() with respect to over the interval [, + h]. 4. Let ƒstd = >t for t Z. a. Find the average rate of change of ƒ with respect to t over the intervals (i) from t = to t = 3, and (ii) from t = to t = T. b. Make a table of values of the average rate of change of ƒ with respect to t over the interval [, T], for some values of T approaching, sa T =.,.,.,.,., and.. c. What does our table indicate is the rate of change of ƒ with respect to t at t =? d. Calculate the it as T approaches of the average rate of change of ƒ with respect to t over the interval from to T.You will have to do some algebra before ou can substitute T =. COMPUTER EXPLORATIONS Graphical Estimates of Limits In Eercises 4 46, use a CAS to perform the following steps: a. Plot the function near the point being approached. b. From our plot guess the value of the it : - : - s + d : : cos : sin : 3-3 cos

14 84 Chapter : Limits and Continuit. HISTORICAL ESSAY* Limits Calculating Limits Using the Limit Laws In Section. we used graphs and calculators to guess the values of its. This section presents theorems for calculating its. The first three let us build on the results of Eample 8 in the preceding section to find its of polnomials, rational functions, and powers. The fourth and fifth prepare for calculations later in the tet. The Limit Laws The net theorem tells how to calculate its of functions that are arithmetic combinations of functions whose its we alread know. THEOREM Limit Laws If L, M, c and k are real numbers and ƒsd = L and gsd = M, then :c :c. Sum Rule: sƒsd + gsdd = L + M :c The it of the sum of two functions is the sum of their its.. Difference Rule: sƒsd - gsdd = L - M :c The it of the difference of two functions is the difference of their its. 3. Product Rule: sƒsd # gsdd = L # M :c The it of a product of two functions is the product of their its. To learn more about the historical figures and the development of the major elements and topics of calculus, visit

15 . Calculating Limits Using the Limit Laws Constant Multiple Rule: sk # ƒsdd = k # L :c The it of a constant times a function is the constant times the it of the function. 5. Quotient Rule: The it of a quotient of two functions is the quotient of their its, provided the it of the denominator is not zero. 6. Power Rule: If r and s are integers with no common factor and s Z, then provided that is a real number. (If s is even, we assume that L 7. ) The it of a rational power of a function is that power of the it of the function, provided the latter is a real number. L r>s :c :c sƒsddr>s = L r>s ƒsd gsd = L M, M Z It is eas to convince ourselves that the properties in Theorem are true (although these intuitive arguments do not constitute proofs). If is sufficientl close to c, then ƒ() is close to L and g() is close to M, from our informal definition of a it. It is then reasonable that ƒsd + gsd is close to L + M; ƒsd - gsd is close to L - M; ƒ()g() is close to LM; kƒ() is close to kl; and that ƒ()>g() is close to L>M if M is not zero. We prove the Sum Rule in Section.3, based on a precise definition of it. Rules 5 are proved in Appendi. Rule 6 is proved in more advanced tets. Here are some eamples of how Theorem can be used to find its of polnomial and rational functions. EXAMPLE Using the Limit Laws Use the observations :c k = k and :c = c (Eample 8 in Section.) and the properties of its to find the following its. (a) (b) (c) : :c s d :c + 5 Solution (a) (b) :c s d = :c :c :c :c + 5 = = = c 3 + 4c - 3 :c s4 + - d :c s + 5d :c :c :c :c + 5 :c = c4 + c - c + 5 Sum and Difference Rules Product and Multiple Rules Quotient Rule Sum and Difference Rules Power or Product Rule

16 86 Chapter : Limits and Continuit (c) : = : - s4-3d Power Rule with r>s = = : : - = 4s -d - 3 Difference Rule Product and Multiple Rules = 6-3 = 3 Two consequences of Theorem further simplif the task of calculating its of polnomials and rational functions. To evaluate the it of a polnomial function as approaches c, merel substitute c for in the formula for the function. To evaluate the it of a rational function as approaches a point c at which the denominator is not zero, substitute c for in the formula for the function. (See Eamples a and b.) THEOREM Limits of Polnomials Can Be Found b Substitution If Psd = a n n + a n - n - + Á + a, then Psd = Pscd = a n :c cn + a n - c n - + Á + a. THEOREM 3 Limits of Rational Functions Can Be Found b Substitution If the Limit of the Denominator Is Not Zero If P() and Q() are polnomials and Qscd Z, then :c Psd Qsd = Pscd Qscd. EXAMPLE Limit of a Rational Function : = s -d3 + 4s -d - 3 s -d + 5 = 6 = This result is similar to the second it in Eample with c = -, now done in one step. Identifing Common Factors It can be shown that if Q() is a polnomial and Qscd =, then s - cd is a factor of Q(). Thus, if the numerator and denominator of a rational function of are both zero at = c, the have s - cd as a common factor. Einating Zero Denominators Algebraicall Theorem 3 applies onl if the denominator of the rational function is not zero at the it point c. If the denominator is zero, canceling common factors in the numerator and denominator ma reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can find the it b substitution in the simplified fraction. EXAMPLE 3 Evaluate Canceling a Common Factor :

17 . Calculating Limits Using the Limit Laws 87 3 (, 3) Solution We cannot substitute = because it makes the denominator zero. We test the numerator to see if it, too, is zero at =. It is, so it has a factor of s - d in common with the denominator. Canceling the s - d s gives a simpler fraction with the same values as the original for Z : = s - ds + d s - d = +, if Z. Using the simpler fraction, we find the it of these values as : b substitution: (a) 3 (, 3) See Figure.8. EXAMPLE 4 Evaluate + - : - + = : = + Creating and Canceling a Common Factor = :. (b) FIGURE.8 The graph of ƒsd = s + - d>s - d in part (a) is the same as the graph of gsd = s + d> in part (b) ecept at =, where ƒ is undefined. The functions have the same it as : (Eample 3). Solution This is the it we considered in Eample of the preceding section. We cannot substitute =, and the numerator and denominator have no obvious common factors. We can create a common factor b multipling both numerator and denominator b the epression + + (obtained b changing the sign after the square root). The preinar algebra rationalizes the numerator: + - = + - # = + - A + + B Therefore, = = A + + B : = : + + = + + Common factor Cancel for Denominator not at ; substitute = =.5. This calculation provides the correct answer to the ambiguous computer results in Eample of the preceding section. The Sandwich Theorem The following theorem will enable us to calculate a variet of its in subsequent chapters. It is called the Sandwich Theorem because it refers to a function ƒ whose values are

18 88 Chapter : Limits and Continuit L h f sandwiched between the values of two other functions g and h that have the same it L at a point c. Being trapped between the values of two functions that approach L, the values of ƒ must also approach L (Figure.9). You will find a proof in Appendi. g c FIGURE.9 The graph of ƒ is sandwiched between the graphs of g and h. THEOREM 4 The Sandwich Theorem Suppose that gsd ƒsd hsd for all in some open interval containing c, ecept possibl at = c itself. Suppose also that Then :c ƒsd = L. gsd = hsd = L. :c :c u() The Sandwich Theorem is sometimes called the Squeeze Theorem or the Pinching Theorem. EXAMPLE 5 Appling the Sandwich Theorem Given that 4 FIGURE. An function u() whose graph lies in the region between = + s >d and = - s >4d has it as : (Eample 5). find : usd, no matter how complicated u is. Solution Since the Sandwich Theorem implies that : usd = (Figure.). EXAMPLE 6-4 usd + for all Z, s - : s >4dd = and s + s >dd =, : More Applications of the Sandwich Theorem (a) (Figure.a). It follows from the definition of sin u that -ƒ u ƒ sin u ƒ u ƒ for all u, and since u: s -ƒ u ƒd = u: ƒ u ƒ =, we have sin u =. u: sin cos (b) (a) FIGURE. The Sandwich Theorem confirms that (a) u: sin u = and (b) u: s - cos ud = (Eample 6).

19 . Calculating Limits Using the Limit Laws 89 (b) (Figure.b). From the definition of cos u, - cos u ƒ u ƒ for all u, and we have u: s - cos ud = or cos u =. u: (c) For an function ƒ(), if :c ƒ ƒsd ƒ =, then :c ƒsd =. The argument: - ƒ ƒsd ƒ ƒsd ƒ ƒsd ƒ and - ƒ ƒsd ƒ and ƒ ƒsd ƒ have it as : c. Another important propert of its is given b the net theorem. A proof is given in the net section. THEOREM 5 If ƒsd gsd for all in some open interval containing c,ecept possibl at = c itself, and the its of ƒ and g both eist as approaches c, then ƒsd gsd. :c :c The assertion resulting from replacing the less than or equal to 6 inequalit in Theorem 5 is false. Figure.a -ƒ u ƒ 6 sin u 6 ƒ u ƒ, but in the it as u :, equalit holds. inequalit b the strict shows that for u Z,

20 . Calculating Limits Using the Limit Laws 89 EXERCISES. Limit Calculations Find the its in Eercises 8.. s + 5d. : : d st - 5dst - 7d 6. t: : : -5. 3s - d. : - 3. s5 - d4>3 4. : h: 3h + + s - 3d : : - s d s:>3 :5 : s + 3d984 : -4 sz - 8d >3 z: h: 3sss - d h + - 5h h: h h: h 5 5h Find the its in Eercises : : -5 t 3. + t - 4. t: t : u u: u : : : : : - t + 3t + t: - t - t - : : : : - + :

21 9 Chapter : Limits and Continuit : -3 Using Limit Rules 37. Suppose : ƒsd = and : gsd = -5. Name the rules in Theorem that are used to accomplish steps (a), (b), and (c) of the following calculation. 38. Let : hsd = 5, : psd =, and : rsd =. Name the rules in Theorem that are used to accomplish steps (a), (b), and (c) of the following calculation. : ƒsd - gsd : sƒsd + 7d = >3 5hsd psds4 - rsdd = = s5ds5d sds4 - d = Suppose :c ƒsd = 5 and :c gsd = -. Find a. ƒsdgsd b. ƒsdgsd :c :c ƒsd c. sƒsd + 3gsdd d. :c :c ƒsd - gsd 4. Suppose :4 ƒsd = and :4 gsd = -3. Find a. sgsd + 3d b. ƒsd :4 :4 gsd c. sgsdd d. :4 :4 ƒsd - 4. Suppose :b ƒsd = 7 and :b gsd = -3. Find a. sƒsd + gsdd b. ƒsd # gsd :b :b c. 4gsd d. ƒsd>gsd :b :b = = = = = : :4 sƒsd - gsdd : sƒsd + 7d >3 : : ƒsd - : gsd A Aƒsd + 7BB >3 : ƒsd - gsd : : A ƒ() + 7B >3 : : sdsd - s -5d = 7 s + 7d >3 4 5hsd : spsds4 - rsddd 5hsd : A p()ba A4 - r()bb : : 5 hsd : A p()ba 4 - r()b : : : (a) (b) (c) (a) (b) (c) 4. Suppose that :- psd = 4, :- rsd =, and :- ssd = -3. Find a. spsd + rsd + ssdd : - b. psd # rsd # ssd : - c. s -4psd + 5rsdd>ssd Limits of Average Rates of Change Because of their connection with secant lines, tangents, and instantaneous rates, its of the form occur frequentl in calculus. In Eercises 43 48, evaluate this it for the given value of and function ƒ. 43. ƒsd =, = 44. ƒsd =, = ƒsd = 3-4, = 46. ƒsd = >, = ƒsd =, = ƒsd = 3 +, = Using the Sandwich Theorem 49. If 5 - ƒsd 5 - for -, find : ƒsd. 5. If - gsd cos for all, find : gsd. 5. a. It can be shown that the inequalities T : - hold for all values of close to zero. What, if anthing, does this tell ou about Give reasons for our answer. b. Graph ƒs + hd - ƒsd h: h sin - cos 6 : = - s >6d, = s sin d>s - cos d, and = together for -. Comment on the behavior of the graphs as :. 5. a. Suppose that the inequalities hold for values of close to zero. (The do, as ou will see in Section.9.) What, if anthing, does this tell ou about Give reasons for our answer. sin - cos? cos 6 - cos :? b. Graph the equations = s>d - s >4d, = s - cos d>, and = > together for -. Comment on the behavior of the graphs as :.

22 . Calculating Limits Using the Limit Laws 9 Theor and Eamples ƒsd a. 53. If 4 ƒsd for in [-, ] and ƒsd for ƒsd b. 4 : - : and 7, at what points c do ou automaticall know 57. a. If :c ƒsd? What can ou sa about the value of the it at ƒsd - 5 = 3, find ƒsd. : - : these points? b. If 54. Suppose that gsd ƒsd hsd for all Z and suppose that ƒsd - 5 = 4, find ƒsd. : - : ƒsd gsd = hsd = If find : : : =, ƒsd Can we conclude anthing about the values of ƒ, g, and h at a. ƒsd b. =? Could ƒsd =? Could : ƒsd =? Give reasons : : for our answers. T 59. a. Graph gsd = sin s>d to estimate : gsd, zooming 55. If ƒsd - 5 in on the origin as necessar. =, find ƒsd. :4 - :4 b. Confirm our estimate in part (a) with a proof. ƒsd T 6. a. Graph hsd = cos s> 3 d to estimate : hsd, zooming 56. If find : - =, in on the origin as necessar. b. Confirm our estimate in part (a) with a proof.

23 .3 The Precise Definition of a Limit 9.3 The Precise Definition of a Limit Now that we have gained some insight into the it concept, working intuitivel with the informal definition, we turn our attention to its precise definition. We replace vague phrases like gets arbitraril close to in the informal definition with specific conditions that can be applied to an particular eample. With a precise definition we will be able to prove conclusivel the it properties given in the preceding section, and we can establish other particular its important to the stud of calculus. To show that the it of ƒ()as : equals the number L,weneed to show that the gap between ƒ() and L can be made as small as we choose if is kept close enough to. Let us see what this would require if we specified the size of the gap between ƒ() and L. 9 To satisf this Restrict to this Upper bound: 9 Lower bound: 5 FIGURE. Keeping within unit of = 4 will keep within units of = 7 (Eample ). EXAMPLE A Linear Function Consider the function = - near = 4. Intuitivel it is clear that is close to 7 when is close to 4, so :4 s - d = 7. However, how close to = 4 does have to be so that = - differs from 7 b, sa, less than units? Solution We are asked: For what values of is ƒ - 7 ƒ 6? To find the answer we first epress ƒ - 7 ƒ in terms of : ƒ - 7 ƒ = ƒ s - d - 7 ƒ = ƒ - 8 ƒ. The question then becomes: what values of satisf the inequalit ƒ - 8 ƒ 6? To find out, we solve the inequalit: ƒ - 8 ƒ Keeping within unit of = 4 will keep within units of = 7 (Figure.).

24 9 Chapter : Limits and Continuit L L L f() f() lies in here for all in here FIGURE.3 How should we define d 7 so that keeping within the interval s - d, + dd will keep ƒ() within the interval al -, L + b? L In the previous eample we determined how close must be to a particular value to ensure that the outputs ƒ() of some function lie within a prescribed interval about a it value L. To show that the it of ƒ() as : actuall equals L, we must be able to show that the gap between ƒ() and L can be made less than an prescribed error, no matter how small, b holding close enough to. Definition of Limit Suppose we are watching the values of a function ƒ() as approaches (without taking on the value of itself). Certainl we want to be able to sa that ƒ() stas within one-tenth of a unit of L as soon as stas within some distance d of (Figure.3). But that in itself is not enough, because as continues on its course toward, what is to prevent ƒ() from jittering about within the interval from L - (>) to L + (>) without tending toward L? We can be told that the error can be no more than > or > or >,. Each time, we find a new d-interval about so that keeping within that interval satisfies the new error tolerance. And each time the possibilit eists that ƒ() jitters awa from L at some stage. The figures on the net page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents P-challenges to prove that the it does not eist or, more precisel, that there is room for doubt, and the scholar answers ever challenge with a d-interval around. How do we stop this seemingl endless series of challenges and responses? B proving that for ever error tolerance P that the challenger can produce, we can find, calculate, or conjure a matching distance d that keeps close enough to to keep ƒ() within that tolerance of L (Figure.4). This leads us to the precise definition of a it. L L f() f() lies in here DEFINITION Limit of a Function Let ƒ() be defined on an open interval about, ecept possibl at itself. We sa that the it of ƒ() as approaches is the number L, and write for all in here ƒsd = L, : if, for ever number P7, there eists a corresponding number d 7 such that for all, 6 ƒ - ƒ 6 d Q ƒ ƒsd - L ƒ 6P. FIGURE.4 The relation of d and P in the definition of it. One wa to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We ma tr for diameter L, but since nothing is perfect, we must be satisfied with a diameter ƒ() somewhere between L -Pand L +P. The d is the measure of how accurate our control setting for must be to guarantee this degree of accurac in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we ma have to adjust d. That is, the value of d, how tight our control setting must be, depends on the value of P, the error tolerance. Eamples: Testing the Definition The formal definition of it does not tell how to find the it of a function, but it enables us to verif that a suspected it is correct. The following eamples show how the

25 .3 The Precise Definition of a Limit 93 f() f() f() f() L L L L L L L L L L L L The challenge: Make f() L / / Response: / (a number) New challenge: Make f() L / / Response: / L f() L f() L L L L New challenge: Response: / L, f() L, f() f() L L, L L, L New challenge:, Response: /, New challenge:... definition can be used to verif it statements for specific functions. (The first two eamples correspond to parts of Eamples 7 and 8 in Section..) However, the real purpose of the definition is not to do calculations like this, but rather to prove general theorems so that the calculation of specific its can be simplified.

26 94 Chapter : Limits and Continuit NOT TO SCALE 5 FIGURE.5 If ƒsd = 5-3, then 6 ƒ - ƒ 6P>5guarantees that ƒ ƒsd - ƒ 6P(Eample ). EXAMPLE Show that Testing the Definition s5-3d =. : Solution Set =, ƒsd = 5-3, and L = in the definition of it. For an given P7, we have to find a suitable d 7 so that if Z and is within distance d of =, that is, whenever 6 ƒ - ƒ 6 d, it is true that ƒ() is within distance P of L =, so ƒ ƒsd - ƒ 6P. We find d b working backward from the P-inequalit: ƒ s5-3d - ƒ = ƒ 5-5 ƒ 6P 5 ƒ - ƒ 6P ƒ - ƒ 6P>5. Thus, we can take d =P>5(Figure.5). If 6 ƒ - ƒ 6 d =P>5, then ƒ s5-3d - ƒ = ƒ 5-5 ƒ = 5 ƒ - ƒ 6 5sP>5d =P, which proves that : s5-3d =. The value of d =P>5is not the onl value that will make 6 ƒ - ƒ 6 d impl ƒ 5-5 ƒ 6P. An smaller positive d will do as well. The definition does not ask for a best positive d, just one that will work. FIGURE.6 For the function ƒsd =, we find that 6 ƒ - ƒ 6 d will guarantee ƒ ƒsd - ƒ 6Pwhenever d P(Eample 3a). EXAMPLE 3 Limits of the Identit and Constant Functions Prove: (a) = (b) k = k (k constant). : : Solution (a) Let P7be given. We must find d 7 such that for all 6 ƒ - ƒ 6 d implies ƒ - ƒ 6P. The implication will hold if d equals P or an smaller positive number (Figure.6). This proves that : =. (b) Let P7be given. We must find d 7 such that for all 6 ƒ - ƒ 6 d implies ƒk - k ƒ 6P. Since k - k =, we can use an positive number for d and the implication will hold (Figure.7). This proves that : k = k. Finding Deltas Algebraicall for Given Epsilons In Eamples and 3, the interval of values about for which ƒ ƒsd - L ƒ was less than P was smmetric about and we could take d to be half the length of that interval. When

27 .3 The Precise Definition of a Limit 95 k k k k to the in- such smmetr is absent, as it usuall is, we can take d to be the distance from terval s nearer endpoint. EXAMPLE 4 Finding Delta Algebraicall For the it :5 - =, find a d 7 that works for P=. That is, find a d 7 such that for all 6 ƒ - 5 ƒ 6 d Q ƒ - - ƒ 6. FIGURE.7 For the function ƒsd = k, we find that ƒ ƒsd - k ƒ 6Pfor an positive d (Eample 3b). Solution We organize the search into two steps, as discussed below.. Solve the inequalit ƒ - - ƒ 6 to find an interval containing = 5 on which the inequalit holds for all Z. ƒ - - ƒ The inequalit holds for all in the open interval (, ), so it holds for all Z 5 in this interval as well (see Figure.9).. Find a value of d 7 to place the centered interval 5 - d d (centered at = 5) inside the interval (, ). The distance from 5 to the nearer endpoint of (, ) is 3 (Figure.8). If we take d = 3 or an smaller positive number, then the inequalit 6 ƒ - 5 ƒ 6 d will automaticall place between and to make ƒ - - ƒ 6 (Figure.9) 6 ƒ - 5 ƒ 6 3 Q ƒ - - ƒ FIGURE.8 An open interval of radius 3 about = 5 will lie inside the open interval (, ) NOT TO SCALE FIGURE.9 The function and intervals in Eample 4.

28 96 Chapter : Limits and Continuit How to Find Algebraicall a D for a Given f, L,, and P> The process of finding a d 7 such that for all 6 ƒ - ƒ 6 d Q ƒƒsd - L ƒ 6P can be accomplished in two steps.. Solve the inequalit ƒ ƒsd - L ƒ 6Pto find an open interval (a, b) containing on which the inequalit holds for all Z.. Find a value of d 7 that places the open interval s - d, + dd centered at inside the interval (a, b). The inequalit ƒ ƒsd - L ƒ 6Pwill hold for all Z in this d-interval (, 4) (, ) 4 FIGURE. An interval containing = so that the function in Eample 5 satisfies ƒ ƒsd - 4 ƒ 6P. EXAMPLE 5 Prove that : ƒsd = 4 if Finding Delta Algebraicall Solution Our task is to show that given P7there eists a d 7 such that for all. Solve the inequalit ƒ ƒsd - 4 ƒ 6Pto find an open interval containing = on which the inequalit holds for all Z. For Z wehave ƒsd =, and the inequalit to solve is ƒ =, - 4 ƒ 6P: The inequalit ƒ ƒsd - 4 ƒ 6Pholds for all Z in the open interval A 4 -P, 4 +PB(Figure.).. Find a value of d 7 that places the centered interval s - d, + dd inside the interval A 4 -P, 4 +PB. Take d to be the distance from = to the nearer endpoint of A 4 -P, 4 +PB. In other words, take d = min E - 4 -P, 4 +P-F, the minimum (the smaller) of the two numbers - 4 -Pand 4 +P-. If d has this or an smaller positive value, the inequalit 6 ƒ - ƒ 6 d will automaticall place between 4 -Pand 4 +Pto make ƒ ƒsd - 4 ƒ 6P. For all, This completes the proof. ƒsd = e, Z, =. 6 ƒ - ƒ 6 d Q ƒƒsd - 4 ƒ 6P. ƒ - 4 ƒ 6P -P 6-4 6P 4 -P P 4 -P6 ƒ ƒ 6 4 +P 4 -P P. 6 ƒ - ƒ 6 d Q ƒƒsd - 4 ƒ 6P. Assumes P64; see below. An open interval about = that solves the inequalit

29 .3 The Precise Definition of a Limit 97 Wh was it all right to assume P64? Because, in finding a d such that for all, 6 ƒ - ƒ 6 d implied ƒ ƒsd - 4 ƒ 6P64, we found a d that would work for an larger P as well. Finall, notice the freedom we gained in letting d = min E - 4 -P, 4 +P- F. We did not have to spend time deciding which, if either, number was the smaller of the two. We just let d represent the smaller and went on to finish the argument. Using the Definition to Prove Theorems We do not usuall rel on the formal definition of it to verif specific its such as those in the preceding eamples. Rather we appeal to general theorems about its, in particular the theorems of Section.. The definition is used to prove these theorems (Appendi ). As an eample, we prove part of Theorem, the Sum Rule. EXAMPLE 6 Proving the Rule for the Limit of a Sum Given that :c ƒsd = L and :c gsd = M, prove that Solution Let P7be given. We want to find a positive number d such that for all 6 ƒ - c ƒ 6 d Q ƒƒsd + gsd - sl + Md ƒ 6P. Regrouping terms, we get sƒsd + gsdd = L + M. :c ƒ ƒsd + gsd - sl + Md ƒ = ƒ sƒsd - Ld + sgsd - Md ƒ ƒ ƒsd - L ƒ + ƒ gsd - M ƒ. Since :c ƒsd = L, there eists a number d 7 such that for all 6 ƒ - c ƒ 6 d Q ƒƒsd - L ƒ 6P>. Similarl, since :c gsd = M, there eists a number d 7 such that for all 6 ƒ - c ƒ 6 d Q ƒgsd - M ƒ 6P>. Let d = min 5d, d 6, the smaller of d and d. If 6 ƒ - c ƒ 6 d then ƒ - c ƒ 6 d, so ƒ ƒsd - L ƒ 6P>, and ƒ - c ƒ 6 d, so ƒ gsd - M ƒ 6P>. Therefore ƒ ƒsd + gsd - sl + Md ƒ 6 P + P =P. This shows that :c sƒsd + gsdd = L + M. Let s also prove Theorem 5 of Section.. Triangle Inequalit: ƒ a + b ƒ ƒ a ƒ + ƒ b ƒ EXAMPLE 7 Given that :c ƒsd = L and :c gsd = M, and that ƒsd gsd for all in an open interval containing c (ecept possibl c itself), prove that L M. Solution We use the method of proof b contradiction. Suppose, on the contrar, that L 7 M. Then b the it of a difference propert in Theorem, s gsd - ƒsdd = M - L. :c Therefore, for an P7, there eists d 7 such that ƒ sgsd - ƒsdd - sm - Ld ƒ 6P whenever 6 ƒ - c ƒ 6 d.

30 98 Chapter : Limits and Continuit Since L - M 7 b hpothesis, we take P=L- M in particular and we have a number d 7 such that ƒ sgsd - ƒsdd - sm - Ld ƒ 6 L - M whenever 6 ƒ - c ƒ 6 d. Since a ƒ a ƒ for an number a, we have sgsd - ƒsdd - sm - Ld 6 L - M whenever 6 ƒ - c ƒ 6 d which simplifies to gsd 6 ƒsd whenever 6 ƒ - c ƒ 6 d. But this contradicts ƒsd gsd. Thus the inequalit L 7 M must be false. Therefore L M.

31 98 Chapter : Limits and Continuit EXERCISES.3 Centering Intervals About a Point In Eercises 6, sketch the interval (a, b) on the -ais with the point inside. Then find a value of d 7 such that for all, 6 ƒ - ƒ 6 d Q a 6 6 b.. a =, b = 7, = 5. a =, b = 7, = 3. a = -7>, b = ->, = a = -7>, b = ->, = -3> 5. a = 4>9, b = 4>7, = > 6. a =.759, b = 3.39, = 3 Finding Deltas Graphicall In Eercises 7 4, use the graphs to find a d 7 such that for all 6 ƒ - ƒ 6 d Q ƒƒsd - L ƒ 6P f() L 7.5 f() NOT TO SCALE 5 L f() L f() L f() 4 L f() 3 L NOT TO SCALE NOT TO SCALE NOT TO SCALE 5 3 NOT TO SCALE

32 .3 The Precise Definition of a Limit f() L Finding Deltas Algebraicall.5 Each of Eercises 5 3 gives a function ƒ() and numbers L, and P7. In each case, find an open interval about on which the inequalit ƒ ƒsd - L ƒ 6Pholds. Then give a value for d 7 such that for all satisfing 6 ƒ - ƒ 6 d the inequalit ƒ ƒsd - L ƒ 6Pholds. 5. ƒsd = +, L = 5, = 4, P =. 6. ƒsd = -, L = -6, = -, P =. 7. ƒsd = +, L =, =, P =. 8. ƒsd =, L = >, = >4, P =. 9. ƒsd = 9 -, L = 3, =, P =. ƒsd = - 7, L = 4, = 3, P =. ƒsd = >, L = >4, = 4, P =.5. ƒsd =, L = 3, = 3, P =. 3. ƒsd =, L = 4, = -, P =.5 4. ƒsd = >, L = -, = -, P =. 5. ƒsd = - 5, L =, = 4, P = 6. ƒsd = >, L = 5, = 4, P = 7. ƒsd = m, m 7, L = m, =, P =.3 8. ƒsd = m, m 7, L = 3m, = 3, P=c7 9. ƒsd = m + b, m 7, L = sm>d + b, = >, P =c7 3. ƒsd = m + b, m 7, L = m + b, =, P= f() L..99 NOT TO SCALE More on Formal Limits Each of Eercises 3 36 gives a function ƒ(), a point, and a positive number P. Find L = ƒsd. Then find a number d 7 such : that for all Prove the it statements in Eercises s9 - d = s3-7d = :4 : = = :9 : = ƒsd = if ƒsd = e, Z :, = ƒsd = 4 if ƒsd = e, Z - : -, = - : = : 3 : -3 6 ƒ - ƒ 6 d Q ƒƒsd - L ƒ 6P. 3. ƒsd = 3 -, = 3, P =. 3. ƒsd = -3 -, = -, P =.3 ƒsd = - 4 -, =, P =.5 ƒsd = , + 5 = -5, P = ƒsd = - 5, = -3, P = ƒsd = 4>, =, P =.4 = 3 4 -, 6 ƒsd = if ƒsd = e : 6-4, Ú, 6 ƒsd = if ƒsd = e : >, Ú : sin = - : - = sin

33 Chapter : Limits and Continuit 5. : sin = volts and I is to be 5 ;. amp. In what interval does R have to lie for I to be within. amp of the value I = 5? V I R sin When Is a Number L Not the Limit of ƒ() as :? We can prove that : ƒsd Z L b providing an P7such that no possible d 7 satisfies the condition For all, 6 ƒ - ƒ 6 d Q ƒƒsd - L ƒ 6P. We accomplish this for our candidate P b showing that for each d 7 there eists a value of such that 6 ƒ - ƒ 6 d and ƒƒsd - L ƒ ÚP. Theor and Eamples 5. Define what it means to sa that gsd = k. 5. Prove that ƒsd = L : if and onl if ƒsh + cd = L. :c h: 53. A wrong statement about its Show b eample that the following statement is wrong. The number L is the it of ƒ() as approaches closer to L as approaches. if ƒ() gets Eplain wh the function in our eample does not have the given value of L as a it as :. 54. Another wrong statement about its Show b eample that the following statement is wrong. The number L is the it of ƒ() as approaches if, given an P7, there eists a value of for which ƒ ƒsd - L ƒ 6P. Eplain wh the function in our eample does not have the given value of L as a it as :. T 55. Grinding engine clinders Before contracting to grind engine clinders to a cross-sectional area of 9 in, ou need to know how much deviation from the ideal clinder diameter of = in. ou can allow and still have the area come within. in of the required 9 in. To find out, ou let A = ps>d and look for the interval in which ou must hold to make ƒ A - 9 ƒ.. What interval do ou find? 56. Manufacturing electrical resistors Ohm s law for electrical circuits like the one shown in the accompaning figure states that V = RI. In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your firm has been asked to suppl the resistors for a circuit in which V will be 57. L Let ƒsd = e, 6 +, 7. L L f() f() f() a value of for which and f() L

34 .3 The Precise Definition of a Limit 58. a. Let P=>. Show that no possible d 7 satisfies the following condition: For all, 6 ƒ - ƒ 6 d Q ƒƒsd - ƒ 6 >. That is, for each d 7 show that there is a value of such that 6 ƒ - ƒ 6 d and ƒƒsd - ƒ Ú >. This will show that : ƒsd Z. b. Show that : ƒsd Z. c. Show that : ƒsd Z.5., 6 Let hsd = 3, =, a. For the function graphed here, show that : - gsd Z. b. Does : - gsd appear to eist? If so, what is the value of the it? If not, wh not? g() 4 3 Show that a. hsd Z 4 : b. hsd Z 3 : c. hsd Z : 59. For the function graphed here, eplain wh a. ƒsd Z 4 :3 b. ƒsd Z 4.8 :3 c. ƒsd Z 3 : h() f() COMPUTER EXPLORATIONS In Eercises 6 66, ou will further eplore finding deltas graphicall. Use a CAS to perform the following steps: a. Plot the function = ƒsd near the point being approached. b. Guess the value of the it L and then evaluate the it smbolicall to see if ou guessed correctl. c. Using the value P=., graph the banding lines = L -P and = L +Ptogether with the function ƒ near. d. From our graph in part (c), estimate a d 7 such that for all 6 ƒ - ƒ 6 d Q ƒƒsd - L ƒ 6P. Test our estimate b plotting ƒ,, and over the interval 6 ƒ - ƒ 6 d. For our viewing window use - d + d and L - P L + P. If an function values lie outside the interval [L -P, L +P], our choice of d was too large. Tr again with a smaller estimate. e. Repeat parts (c) and (d) successivel for P=.,.5, and.. 6. ƒsd = , = 3 6. ƒsd = , = sin 63. ƒsd = 3, = s - cos d 64. ƒsd = - sin, = 65. ƒsd = 3 - -, = 66. ƒsd = 3 - s7 + d + 5, - =

35 Chapter : Limits and Continuit.4 One-Sided Limits and Limits at Infinit In this section we etend the it concept to one-sided its, which are its as approaches the number from the left-hand side (where 6 ) or the right-hand side s 7 d onl. We also analze the graphs of certain rational functions as well as other functions with it behavior as : ; q. FIGURE. Different right-hand and left-hand its at the origin. One-Sided Limits To have a it L as approaches c, a function ƒ must be defined on both sides of c and its values ƒ() must approach L as approaches c from either side. Because of this, ordinar its are called two-sided. If ƒ fails to have a two-sided it at c, it ma still have a one-sided it, that is, a it if the approach is onl from one side. If the approach is from the right, the it is a right-hand it. From the left, it is a left-hand it. The function ƒsd = > ƒ ƒ (Figure.) has it as approaches from the right, and it - as approaches from the left. Since these one-sided it values are not the same, there is no single number that ƒ() approaches as approaches. So ƒ() does not have a (two-sided) it at. Intuitivel, if ƒ() is defined on an interval (c, b), where c 6 b, and approaches arbitraril close to L as approaches c from within that interval, then ƒ has right-hand it L at c. We write The smbol : c + means that we consider onl values of greater than c. Similarl, if ƒ() is defined on an interval (a, c), where a 6 c and approaches arbitraril close to M as approaches c from within that interval, then ƒ has left-hand it M at c. We write The smbol : c - means that we consider onl values less than c. These informal definitions are illustrated in Figure.. For the function ƒsd = > ƒ ƒ in Figure. we have : ƒsd = L. + :c ƒsd = M. - :c ƒsd = and ƒsd = : L f() f() M c c (a) f() L c (b) f() M c FIGURE. (a) Right-hand it as approaches c. (b) Left-hand it as approaches c.

36 .4 One-Sided Limits and Limits at Infinit 3 4 FIGURE.3 and : = - (Eample ). : - = + EXAMPLE One-Sided Limits for a Semicircle The domain of ƒsd = 4 - is [-, ]; its graph is the semicircle in Figure.3. We have 4 - : - = and : =. - The function does not have a left-hand it at = - or a right-hand it at =. It does not have ordinar two-sided its at either - or. One-sided its have all the properties listed in Theorem in Section.. The righthand it of the sum of two functions is the sum of their right-hand its, and so on. The theorems for its of polnomials and rational functions hold with one-sided its, as does the Sandwich Theorem and Theorem 5. One-sided its are related to its in the following wa. THEOREM 6 A function ƒ() has a it as approaches c if and onl if it has left-hand and right-hand its there and these one-sided its are equal: :c ƒsd = L 3 :c ƒsd = L and ƒsd = L. - + :c FIGURE.4 in Eample. f() 3 4 Graph of the function EXAMPLE Limits of the Function Graphed in Figure.4 At = : : + ƒsd =, : - ƒsd and : ƒsd do not eist. The function is not defined to the left of =. At = : : - ƒsd = even though ƒsd =, : + ƒsd =, : ƒsd does not eist. The right- and left-hand its are not equal. At = : : - ƒsd =, : + ƒsd =, : ƒsd = even though ƒsd =. At = 3: :3 - ƒsd = :3 + ƒsd = :3 ƒsd = ƒs3d =. At = 4: :4 - ƒsd = even though ƒs4d Z, :4 + ƒsd and :4 ƒsd do not eist. The function is not defined to the right of = 4. At ever other point c in [, 4], ƒ() has it ƒ(c). Precise Definitions of One-Sided Limits The formal definition of the it in Section.3 is readil modified for one-sided its.

37 4 Chapter : Limits and Continuit DEFINITIONS Right-Hand, Left-Hand Limits We sa that ƒ() has right-hand it L at, and write L L f() f() lies in here ƒsd = L (See Figure.5) + : if for ever number P7there eists a corresponding number d 7 such that for all d Q ƒƒsd - L ƒ 6P. L We sa that ƒ has left-hand it L at, and write for all in here ƒsd = L (See Figure.6) - : if for ever number P7there eists a corresponding number d 7 such that for all - d 6 6 Q ƒƒsd - L ƒ 6P. FIGURE.5 Intervals associated with the definition of right-hand it. EXAMPLE 3 Prove that Appling the Definition to Find Delta =. + : L L L f() f() lies in here for all in here FIGURE.6 Intervals associated with the definition of left-hand it. Solution Let P7be given. Here = and L =, so we want to find a d 7 such that for all 6 6 d Q ƒ - ƒ 6P, or 6 6 d Q 6P. Squaring both sides of this last inequalit gives 6P if 6 6 d. If we choose d =P we have 6 6 d =P Q 6P, or 6 6P Q ƒ - ƒ 6P. According to the definition, this shows that : + = (Figure.7). f() f() L FIGURE.7 = in Eample 3. : + The functions eamined so far have had some kind of it at each point of interest. In general, that need not be the case. EXAMPLE 4 A Function Oscillating Too Much Show that = sin s>d has no it as approaches zero from either side (Figure.8). Solution As approaches zero, its reciprocal, >, grows without bound and the values of sin ( >) ccle repeatedl from - to. There is no single number L that the function s

38 .4 One-Sided Limits and Limits at Infinit 5 sin FIGURE.8 The function = sin s>d has neither a right-hand nor a left-hand it as approaches zero (Eample 4). values sta increasingl close to as approaches zero. This is true even if we restrict to positive values or to negative values. The function has neither a right-hand it nor a lefthand it at =. Limits Involving (sin U)/U A central fact about ssin ud>u is that in radian measure its it as u : is. We can see this in Figure.9 and confirm it algebraicall using the Sandwich Theorem. sin (radians) 3 3 NOT TO SCALE T FIGURE.9 The graph of ƒsud = ssin ud>u. P sin tan THEOREM 7 sin u u: u = su in radiansd () cos O Q A(, ) FIGURE.3 The figure for the proof of Theorem 7. TA>OA = tan u, but OA =, so TA = tan u. Proof The plan is to show that the right-hand and left-hand its are both. Then we will know that the two-sided it is as well. To show that the right-hand it is, we begin with positive values of u less than p> (Figure.3). Notice that Area OAP 6 area sector OAP 6 area OAT.

39 6 Chapter : Limits and Continuit Equation () is where radian measure comes in: The area of sector OAP is u> onl if u is measured in radians. We can epress these areas in terms of u as follows: Area OAP = base * height = sdssin ud = sin u Area sector OAP = r u = sd u = u () Area OAT = base * height = sdstan ud = tan u. Thus, This last inequalit goes the same wa if we divide all three terms b the number ( >) sin u, which is positive since 6 u 6 p>: Taking reciprocals reverses the inequalities: Since u: + cos u = (Eample 6b, Section.), the Sandwich Theorem gives Recall that sin u and u are both odd functions (Section.4). Therefore, ƒsud = ssin ud>u is an even function, with a graph smmetric about the -ais (see Figure.9). This smmetr implies that the left-hand it at eists and has the same value as the right-hand it: sin u u: - u so u: ssin ud>u = b Theorem 6. sin u 6 u 6 tan u. 6 u sin u 6 cos u. 7 sin u u sin u u: + u 7 cos u. =. = = u: + sin u u, EXAMPLE 5 Using sin u u: u = Show that (a) cos h - h: h = and (b) sin : 5 = 5. Solution (a) Using the half-angle formula cos h = - sin sh>d, we calculate cos h - h: h = - sin sh>d h: h sin u = - u: u sin u = -sdsd =. Let u = h>.

40 .4 One-Sided Limits and Limits at Infinit 7 (b) Equation () does not appl to the original fraction. We need a in the denominator, not a 5. We produce it b multipling numerator and denominator b >5: 4 sin : 5 = : s>5d # sin s>5d # 5 = 5 sin : Now, Eq. () applies with u. 3 = 5 sd = 5 FIGURE The graph of = >. Finite Limits as : ˆ The smbol for infinit s q d does not represent a real number. We use q to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For eample, the function ƒsd = > is defined for all Z (Figure.3). When is positive and becomes increasingl large, > becomes increasingl small. When is negative and its magnitude becomes increasingl large, > again becomes small. We summarize these observations b saing that ƒsd = > has it as : ; q or that is a it of ƒsd = > at infinit and negative infinit. Here is a precise definition. DEFINITIONS Limit as approaches ˆ or ˆ. We sa that ƒ() has the it L as approaches infinit and write ƒsd = L : q if, for ever number P7, there eists a corresponding number M such that for all 7 M Q ƒƒsd - L ƒ 6P.. We sa that ƒ() has the it L as approaches minus infinit and write ƒsd = L : -q if, for ever number P7, there eists a corresponding number N such that for all 6 N Q ƒƒsd - L ƒ 6P. Intuitivel, : q ƒsd = L if, as moves increasingl far from the origin in the positive direction, ƒ() gets arbitraril close to L. Similarl, :- q ƒsd = L if, as moves increasingl far from the origin in the negative direction, ƒ() gets arbitraril close to L. The strateg for calculating its of functions as : ; q is similar to the one for finite its in Section.. There we first found the its of the constant and identit functions = k and =. We then etended these results to other functions b appling a theorem about its of algebraic combinations. Here we do the same thing, ecept that the starting functions are = k and = > instead of = k and =.

41 8 Chapter : Limits and Continuit N No matter what positive number is, the graph enters this band at and stas. No matter what positive number is, the graph enters this band at and stas. M The basic facts to be verified b appling the formal definition are We prove the latter and leave the former to Eercises 7 and 7. EXAMPLE 6 Show that (a) : q = k = k and : ;q : ;q Limits at Infinit for ƒsd = Solution (a) Let P7be given. We must find a number M such that for all (b) : -q =. =. (3) FIGURE.3 The geometr behind the argument in Eample 6. 7 M Q ` - ` = ` ` 6P. The implication will hold if M = >P or an larger positive number (Figure.3). This proves : q s>d =. (b) Let P7be given. We must find a number N such that for all 6 N Q ` - ` = ` ` 6P. The implication will hold if N = ->P or an number less than ->P (Figure.3). This proves :- q s>d =. Limits at infinit have properties similar to those of finite its. THEOREM 8 Limit Laws as : ˆ If L, M, and k, are real numbers and. Sum Rule:. Difference Rule: ƒsd = L and gsd = M, then : ;q : ;q sƒsd + gsdd = L + M : ;q : ;q sƒsd - gsdd = L - M 3. Product Rule: sƒsd # gsdd = L # M : ;q 4. Constant Multiple Rule: sk # ƒsdd = k # L : ;q ƒsd 5. Quotient Rule: : ;q gsd = L M, M Z 6. Power Rule: If r and s are integers with no common factors, s Z, then : ;q sƒsddr>s = L r>s provided that L r>s is a real number. (If s is even, we assume that L 7. )

42 .4 One-Sided Limits and Limits at Infinit 9 These properties are just like the properties in Theorem, Section., and we use them the same wa. EXAMPLE 7 Using Theorem Line 5 3 NOT TO SCALE FIGURE.33 The graph of the function in Eample 8. The graph approaches the line = 5>3 as ƒ ƒ increases (a) (b) a5 + : q b = 5 + : q : q p3 : -q = p3 # # : -q Sum Rule Known its Product rule Known its Limits at Infinit of Rational Functions To determine the it of a rational function as : ; q, we can divide the numerator and denominator b the highest power of in the denominator. What happens then depends on the degrees of the polnomials involved. EXAMPLE 8 EXAMPLE 9 = 5 + = 5 = : -q p3 # : -q = p3 # # = Numerator and Denominator of Same Degree : q 3 + Degree of Numerator Less Than Degree of Denominator + : -q 3 - = # : -q 5 + s8>d - s3> d = : q 3 + s> d = s> d + s> 3 d : -q - s> 3 d = + - = = 5 3 Divide numerator and denominator b. See Fig..33. Divide numerator and denominator b 3. See Fig..34. ƒ ƒ FIGURE.34 The graph of the function in Eample 9. The graph approaches the -ais as increases. We give an eample of the case when the degree of the numerator is greater than the degree of the denominator in the net section (Eample 8, Section.5). Horizontal Asmptotes If the distance between the graph of a function and some fied line approaches zero as a point on the graph moves increasingl far from the origin, we sa that the graph approaches the line asmptoticall and that the line is an asmptote of the graph. Looking at ƒsd = > (See Figure.3), we observe that the -ais is an asmptote of the curve on the right because and on the left because : q = : -q =.

43 Chapter : Limits and Continuit We sa that the -ais is a horizontal asmptote of the graph of ƒsd = >. DEFINITION Horizontal Asmptote A line = b is a horizontal asmptote of the graph of a function = ƒsd if either ƒsd = b or ƒsd = b. : q : -q The curve sketched in Figure.33 (Eample 8) has the line = 5>3 as a horizontal asmptote on both the right and the left because EXAMPLE Find sin s>d. : q ƒsd = ƒsd = 5 and : q 3 ƒsd = 5 : -q 3. Substituting a New Variable Solution We introduce the new variable t = >. From Eample 6, we know that t : + as : q (see Figure.3). Therefore, sin : q = sin t =. + t: sin The Sandwich Theorem Revisited The Sandwich Theorem also holds for its as : ; q. EXAMPLE A Curve Ma Cross Its Horizontal Asmptote Using the Sandwich Theorem, find the horizontal asmptote of the curve = + sin. Solution We are interested in the behavior as : ; q. Since sin ` ` ` ` 3 3 FIGURE.35 A curve ma cross one of its asmptotes infinitel often (Eample ). and :; q ƒ > ƒ =, we have :; q ssin d> = b the Sandwich Theorem. Hence, sin a + : ;q b = + =, and the line = is a horizontal asmptote of the curve on both left and right (Figure.35).

44 .4 One-Sided Limits and Limits at Infinit This eample illustrates that a curve ma cross one of its horizontal asmptotes, perhaps man times. Oblique Asmptotes If the degree of the numerator of a rational function is one greater than the degree of the denominator, the graph has an oblique (slanted) asmptote. We find an equation for the asmptote b dividing numerator b denominator to epress ƒ as a linear function plus a remainder that goes to zero as : ; q. Here s an eample. EXAMPLE Finding an Oblique Asmptote Find the oblique asmptote for the graph of ƒsd = in Figure Solution B long division, we find ƒsd = = a b s7 + 4d ('')''* ('')''* linear function gsd remainder As : ; q, the remainder, whose magnitude gives the vertical distance between the graphs of ƒ and g, goes to zero, making the (slanted) line 4 FIGURE.36 The function in Eample has an oblique asmptote. gsd = an asmptote of the graph of ƒ (Figure.36). The line = gsd is an asmptote both to the right and to the left. In the net section ou will see that the function ƒ() grows arbitraril large in absolute value as approaches -4>7, where the denominator becomes zero (Figure.36).

45 .5 Infinite Limits and Vertical Asmptotes 5.5 Infinite Limits and Vertical Asmptotes In this section we etend the concept of it to infinite its, which are not its as before, but rather an entirel new use of the term it. Infinite its provide useful smbols and language for describing the behavior of functions whose values become arbitraril large, positive or negative. We continue our analsis of graphs of rational functions from the last section, using vertical asmptotes and dominant terms for numericall large values of. You can get as low as ou want b taking close enough to. FIGURE.37 B You can get as high as ou want b taking close enough to. No matter how high B is, the graph goes higher. B No matter how low B is, the graph goes lower. One-sided infinite its: : + = q and : - = -q Infinite Limits Let us look again at the function ƒsd = >. As : +, the values of ƒ grow without bound, eventuall reaching and surpassing ever positive real number. That is, given an positive real number B, however large, the values of ƒ become larger still (Figure.37). Thus, ƒ has no it as : +. It is nevertheless convenient to describe the behavior of ƒ b saing that ƒ() approaches q as : +. We write ƒsd = + : : = q. + In writing this, we are not saing that the it eists. Nor are we saing that there is a real number q, for there is no such number. Rather, we are saing that : + s>d does not eist because > becomes arbitraril large and positive as : +. As : -, the values of ƒsd = > become arbitraril large and negative. Given an negative real number -B, the values of ƒ eventuall lie below -B. (See Figure.37.) We write ƒsd = - : : = -q. - Again, we are not saing that the it eists and equals the number - q. There is no real number - q. We are describing the behavior of a function whose it as : - does not eist because its values become arbitraril large and negative. EXAMPLE One-Sided Infinite Limits 3 Find : + - and : - -. Geometric Solution The graph of = >s - d is the graph of = > shifted unit to the right (Figure.38). Therefore, = >s - d behaves near eactl the wa = > behaves near : FIGURE.38 Near =, the function = >s - d behaves the wa the function = > behaves near =. Its graph is the graph of = > shifted unit to the right (Eample ). : + - = q and : - - = - q. Analtic Solution Think about the number - and its reciprocal. As : +, we have s - d : + and >s - d : q. As : -, we have s - d : - and >s - d : - q.

46 6 Chapter : Limits and Continuit B No matter how high B is, the graph goes higher. EXAMPLE Discuss the behavior of (a) ƒsd = near =, Two-Sided Infinite Limits f() (b) gsd = near = -3. s + 3d (a) Solution (a) As approaches zero from either side, the values of > are positive and become arbitraril large (Figure.39a): g() ( 3) (b) FIGURE.39 The graphs of the functions in Eample. (a) ƒ() approaches infinit as :. (b) g() approaches infinit as : -3. (b) The graph of gsd = >s + 3d is the graph of ƒsd = > shifted 3 units to the left (Figure.39b). Therefore, g behaves near -3 eactl the wa ƒ behaves near. The function = > shows no consistent behavior as :. We have > : q if : +, but > : - q if : -. All we can sa about : s>d is that it does not eist. The function = > is different. Its values approach infinit as approaches zero from either side, so we can sa that : s> d = q. EXAMPLE 3 (a) (b) (c) (d) (e) : : - 3 : = - 3 : = : s - d - 4 Rational Functions Can Behave in Various Was Near Zeros of Their Denominators = : = : = : ƒsd = : : = q. gsd = : -3 : -3 s + 3d = q. s - d s - ds + d = : - s - ds + d = : - 3 : + s - ds + d = - q - 3 : - s - ds + d = q - 3 s - ds + d + = 4 does not eist. - + = The values are negative for 7, near. The values are positive for 6, near. See parts (c) and (d). - (f) : s - d 3 = -s - d - : s - d 3 = : s - d = - q In parts (a) and (b) the effect of the zero in the denominator at = is canceled because the numerator is zero there also. Thus a finite it eists. This is not true in part (f), where cancellation still leaves a zero in the denominator. Precise Definitions of Infinite Limits Instead of requiring ƒ() to lie arbitraril close to a finite number L for all sufficientl close to, the definitions of infinite its require ƒ() to lie arbitraril far from the ori-

47 .5 Infinite Limits and Vertical Asmptotes 7 f() gin. Ecept for this change, the language is identical with what we have seen before. Figures.4 and.4 accompan these definitions. DEFINITIONS Infinit, Negative Infinit as Limits B. We sa that ƒ() approaches infinit as approaches, and write ƒsd = q, : if for ever positive real number B there eists a corresponding d 7 such that for all 6 ƒ - ƒ 6 d Q ƒsd 7 B. FIGURE.4 :. ƒ() approaches infinit as. We sa that ƒ() approaches negative infinit as approaches, and write ƒsd = -q, : if for ever negative real number -B there eists a corresponding d 7 such that for all 6 ƒ - ƒ 6 d Q ƒsd 6 -B. The precise definitions of one-sided infinite its at eercises. are similar and are stated in the B EXAMPLE 4 Using the Definition of Infinite Limits f() Prove that : = q. Solution Given B 7, we want to find d 7 such that FIGURE.4 ƒ() approaches negative infinit as :. Now, or, equivalentl, 6 ƒ - ƒ 6 d implies 7 B. 7 B if and onl if 6 B ƒ ƒ 6 B. Thus, choosing d = > B (or an smaller positive number), we see that ƒ ƒ 6 d implies 7 d Ú B. Therefore, b definition, : = q.

48 8 Chapter : Limits and Continuit Vertical asmptote Vertical Asmptotes Notice that the distance between a point on the graph of = > and the -ais approaches zero as the point moves verticall along the graph and awa from the origin (Figure.4). This behavior occurs because Horizontal asmptote Horizontal asmptote, : + = q and : - = -q. We sa that the line = (the -ais) is a vertical asmptote of the graph of = >. Observe that the denominator is zero at = and the function is undefined there. Vertical asmptote, FIGURE.4 The coordinate aes are asmptotes of both branches of the hperbola = >. Horizontal asmptote, Vertical asmptote, FIGURE.43 The lines = and = - are asmptotes of the curve = s + 3d>s + d (Eample 5). DEFINITION Vertical Asmptote A line = a is a vertical asmptote of the graph of a function = ƒsd if either ƒsd = ; q or ƒsd = ; q. :a + :a - EXAMPLE 5 Looking for Asmptotes Find the horizontal and vertical asmptotes of the curve Solution We are interested in the behavior as : ; q and as : -, where the denominator is zero. The asmptotes are quickl revealed if we recast the rational function as a polnomial with a remainder, b dividing s + d into s + 3d. This result enables us to rewrite : We now see that the curve in question is the graph of = > shifted unit up and units left (Figure.43). The asmptotes, instead of being the coordinate aes, are now the lines = and = -. EXAMPLE 6 = = + Asmptotes Need Not Be Two-Sided Find the horizontal and vertical asmptotes of the graph of ƒsd = Solution We are interested in the behavior as : ; q and as : ;, where the denominator is zero. Notice that ƒ is an even function of, so its graph is smmetric with respect to the -ais. (a) The behavior as : ; q. Since :q ƒsd =, the line = is a horizontal asmptote of the graph to the right. B smmetr it is an asmptote to the left as well

49 .5 Infinite Limits and Vertical Asmptotes 9 Vertical asmptote, Vertical asmptote, Horizontal asmptote, 3 4 FIGURE.44 Graph of = -8>s - 4d. Notice that the curve approaches the -ais from onl one side. Asmptotes do not have to be two-sided (Eample 6). (Figure.44). Notice that the curve approaches the -ais from onl the negative side (or from below). (b) The behavior as : ;. Since the line = is a vertical asmptote both from the right and from the left. B smmetr, the same holds for the line = -. There are no other asmptotes because ƒ has a finite it at ever other point. EXAMPLE 7 The curves ƒsd = - q and ƒsd = q, : + : - Curves with Infinitel Man Asmptotes = sec = cos and = tan = cos sin both have vertical asmptotes at odd-integer multiples of p>, where cos = (Figure.45). sec tan FIGURE.45 The graphs of sec and tan have infinitel man vertical asmptotes (Eample 7). The graphs of = csc = and = cot = cos sin sin have vertical asmptotes at integer multiples of p, where sin = (Figure.46). csc cot 3 3 FIGURE.46 The graphs of csc and cot (Eample 7).

50 Chapter : Limits and Continuit EXAMPLE 8 A Rational Function with Degree of Numerator Greater than Degree of Denominator Find the asmptotes of the graph of Vertical asmptote, 3 The vertical distance between curve and line goes to zero as Oblique asmptote FIGURE.47 The graph of ƒsd = s - 3d>s - 4d has a vertical asmptote and an oblique asmptote (Eample 8). Solution We are interested in the behavior as : ; q and also as :, where the denominator is zero. We divide s - 4d into s - 3d: This tells us that ƒsd = ƒsd = = linear remainder Since : + ƒsd = q and : - ƒsd = -q, the line = is a two-sided vertical asmptote. As : ; q, the remainder approaches and ƒsd : s>d +. The line = s>d + is an oblique asmptote both to the right and to the left (Figure.47). Notice in Eample 8, that if the degree of the numerator in a rational function is greater than the degree of the denominator, then the it is + q or - q, depending on the signs assumed b the numerator and denominator as ƒ ƒ becomes large. Dominant Terms Of all the observations we can make quickl about the function in Eample 8, probabl the most useful is that This tells us immediatel that ƒsd = ƒsd = ƒsd L + ƒsd L - 4 For numericall large For near If we want to know how ƒ behaves, this is the wa to find out. It behaves like = s>d + when is numericall large and the contribution of >s - 4d to the total

51 .5 Infinite Limits and Vertical Asmptotes value of ƒ is insignificant. It behaves like >s - 4d when is so close to that >s - 4d makes the dominant contribution. We sa that s>d + dominates when is numericall large, and we sa that >s - 4d dominates when is near. Dominant terms like these are the ke to predicting a function s behavior. Here s another eample. EXAMPLE 9 Two Graphs Appearing Identical on a Large Scale Let ƒsd = and gsd = 3 4. Show that although ƒ and g are quite different for numericall small values of, the are virtuall identical for ƒ ƒ ver large. Solution The graphs of ƒ and g behave quite differentl near the origin (Figure.48a), but appear as virtuall identical on a larger scale (Figure.48b). 5, 5 3, 5 f() g(), 5, (a) (b) FIGURE.48 The graphs of ƒ and g, (a) are distinct for ƒ ƒ small, and (b) nearl identical for ƒ ƒ large (Eample 9). We can test that the term 3 4 in ƒ, represented graphicall b g, dominates the polnomial ƒ for numericall large values of b eamining the ratio of the two functions as : ; q. We find that ƒsd : ;q gsd = so that ƒ and g are nearl identical for ƒ ƒ large : ;q 3 4 = a - : ;q b =,

52 Chapter : Limits and Continuit EXERCISES.5 Infinite Limits Find the its in Eercises... : :7 s - 7d 9. a. b.. a. b. 4.. Find the its in Eercises tan 4. :sp>d - 5. s + csc ud 6. - Additional Calculations Find the its in Eercises : - 3 : -8 + : u: : + 3 >3 : + >5 >5-4 as a. : + b. : - c. : - + d. : as a. : + b. : - c. : - + d. : - - a - b as a. : + b. : - c. : 3 d. : as : - 5 : : : : - s + d : - 3 >3 : - >5 >3 sec :s-p>d + s - cot ud u: a. : - + b. : - - c. : + d. : as a. : + b. : + c. : - d. : e. What, if anthing, can be said about the it as :? as a. : + b. : - + c. : - d. : + e. What, if anthing, can be said about the it as :? Find the its in Eercises a - 3 b as >3 t a. t : + b. t : - 4. a + 7b as 3>5 t a. t : + b. t : - 5. a + b as >3 >3 s - d a. : + b. : - c. : + d. : - 6. a - b as >3 4>3 s - d a. : + b. : - c. : + d. : - Graphing Rational Functions Graph the rational functions in Eercises Include the graphs and equations of the asmptotes and dominant terms. 7. = 8. = = 3. = = = = + = = - = = = 3 +

53 .5 Infinite Limits and Vertical Asmptotes 3 Inventing Graphs from Values and Limits In Eercises 39 4, sketch the graph of a function = ƒsd that satisfies the given conditions. No formulas are required just label the coordinate aes and sketch an appropriate graph. (The answers are not unique, so our graphs ma not be eactl like those in the answer section.) 39. ƒsd =, ƒsd =, ƒs -d = -, ƒsd = -, and ƒsd = : -q : q 4. ƒsd =, ƒsd =, ƒsd =, and + : ;q : ƒsd = - - : 4. ƒsd =, ƒsd =, ƒsd = ƒsd = q, - + : ;q : : - ƒsd = -q, and ƒsd = -q + - : : - 4. ƒsd =, ƒs -d =, ƒsd =, ƒsd = q, : q + ƒsd = -q, and ƒsd = : - : : -q Inventing Functions In Eercises 43 46, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. An function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) 43. ƒsd =, ƒsd = q, and ƒsd = q - : ;q : : gsd =, gsd = -q, and gsd = q - : ;q :3 : hsd = -, hsd =, hsd = -, and : -q hsd = : q - : : ksd =, ksd = q, and ksd = -q - + : ;q : : The Formal Definition of Infinite Limit ƒ ƒ Use formal definitions to prove the it statements in Eercises = q : = -q : : -5 s + 5d = q :3 s - 3d = -q Formal Definitions of Infinite One-Sided Limits 5. Here is the definition of infinite right-hand it. We sa that ƒ() approaches infinit as approaches from the right, and write ƒsd = q, + : if, for ever positive real number B, there eists a corresponding number d 7 such that for all d Q ƒsd 7 B. Modif the definition to cover the following cases. a. ƒsd = q - : b. ƒsd = -q + : c. ƒsd = -q Use the formal definitions from Eercise 5 to prove the it statements in Eercises Graphing Terms Each of the functions in Eercises 57 6 is given as the sum or difference of two terms. First graph the terms (with the same set of aes). Then, using these graphs as guides, sketch in the graph of the function : - : = q + : = -q - : - - = -q : + - = q : - - = q = sec +, - p 6 6 p = sec -, - p 6 6 p = tan +, - p 6 6 p = - tan, - p 6 6 p Grapher Eplorations Comparing Graphs with Formulas Graph the curves in Eercises Eplain the relation between the curve s formula and what ou see. 6. = 4-6. = = >3 + >3 p 64. = sin a + b

54 4 Chapter : Limits and Continuit.6 Continuit Distance fallen (m) Q Q Elapsed time (sec) Q 3 Q 4 5 FIGURE.49 Connecting plotted points b an unbroken curve from eperimental data Q, Q, Q 3, Á for a falling object. Continuit from the right Two-sided continuit f() a c b P Continuit from the left FIGURE.5 Continuit at points a, b, and c. 3 f() FIGURE.5 The function is continuous on [, 4] ecept at =, =, and = 4 (Eample ). 4 t When we plot function values generated in a laborator or collected in the field, we often connect the plotted points with an unbroken curve to show what the function s values are likel to have been at the times we did not measure (Figure.49). In doing so, we are assuming that we are working with a continuous function, so its outputs var continuousl with the inputs and do not jump from one value to another without taking on the values in between. The it of a continuous function as approaches c can be found simpl b calculating the value of the function at c. (We found this to be true for polnomials in Section..) An function = ƒsd whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an eample of a continuous function. In this section we investigate more precisel what it means for a function to be continuous. We also stud the properties of continuous functions, and see that man of the function tpes presented in Section.4 are continuous. Continuit at a Point To understand continuit, we need to consider a function like the one in Figure.5 whose its we investigated in Eample, Section.4. EXAMPLE Investigating Continuit Find the points at which the function ƒ in Figure.5 is continuous and the points at which ƒ is discontinuous. Solution The function ƒ is continuous at ever point in its domain [, 4] ecept at =, =, and = 4. At these points, there are breaks in the graph. Note the relationship between the it of ƒ and the value of ƒ at each point of the function s domain. Points at which ƒ is continuous: At =, At = 3, At 6 c 6 4, c Z,, Points at which ƒ is discontinuous: ƒsd = ƒsd. + : ƒsd = ƒs3d. :3 ƒsd = ƒscd. :c At does not eist. : At ƒsd =, but Z ƒsd. : At ƒsd =, but Z ƒs4d. - :4 At c 6, c 7 4, these points are not in the domain of ƒ. To define continuit at a point in a function s domain, we need to define continuit at an interior point (which involves a two-sided it) and continuit at an endpoint (which involves a one-sided it) (Figure.5).

55 .6 Continuit 5 DEFINITION Continuous at a Point Interior point: A function = ƒsd is continuous at an interior point c of its domain if Endpoint: A function = ƒsd is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if :a ƒsd = ƒscd. :c ƒsd = ƒsad or ƒsd = ƒsbd, respectivel. + - :b 4 FIGURE.5 A function that is continuous at ever domain point (Eample ). If a function ƒ is not continuous at a point c, we sa that ƒ is discontinuous at c and c is a point of discontinuit of ƒ. Note that c need not be in the domain of ƒ. A function ƒ is right-continuous (continuous from the right) at a point = c in its domain if :c + ƒsd = ƒscd. It is left-continuous (continuous from the left) at c if :c - ƒsd = ƒscd. Thus, a function is continuous at a left endpoint a of its domain if it is right-continuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its domain if and onl if it is both right-continuous and left-continuous at c (Figure.5). EXAMPLE A Function Continuous Throughout Its Domain The function ƒsd = 4 - is continuous at ever point of its domain, [-, ] (Figure.5), including = -, where ƒ is right-continuous, and =, where ƒ is left-continuous. U() FIGURE.53 A function that is right-continuous, but not left-continuous, at the origin. It has a jump discontinuit there (Eample 3). EXAMPLE 3 The Unit Step Function Has a Jump Discontinuit The unit step function U(), graphed in Figure.53, is right-continuous at =, but is neither left-continuous nor continuous there. It has a jump discontinuit at =. We summarize continuit at a point in the form of a test. Continuit Test A function ƒ() is continuous at = c if and onl if it meets the following three conditions.. ƒ(c) eists (c lies in the domain of ƒ). :c ƒsd eists (ƒ has a it as : c) 3. :c ƒsd = ƒscd (the it equals the function value) For one-sided continuit and continuit at an endpoint, the its in parts and 3 of the test should be replaced b the appropriate one-sided its.

56 6 Chapter : Limits and Continuit 4 3 int or FIGURE.54 The greatest integer function is continuous at ever noninteger point. It is right-continuous, but not left-continuous, at ever integer point (Eample 4). 3 4 EXAMPLE 4 The Greatest Integer Function The function = :; or = int, introduced in Chapter, is graphed in Figure.54. It is discontinuous at ever integer because the it does not eist at an integer n: int = n - and int = n - :n :n + so the left-hand and right-hand its are not equal as : n. Since int n = n, the greatest integer function is right-continuous at ever integer n (but not left-continuous). The greatest integer function is continuous at ever real number other than the integers. For eample, int = = int.5. :.5 In general, if n - 6 c 6 n, n an integer, then int = n - = int c. :c Figure.55 is a catalog of discontinuit tpes. The function in Figure.55a is continuous at =. The function in Figure.55b would be continuous if it had ƒsd =. The function in Figure.55c would be continuous if ƒ() were instead of. The discontinuities in Figure.55b and c are removable. Each function has a it as :, and we can remove the discontinuit b setting ƒ() equal to this it. The discontinuities in Figure.55d through f are more serious: : ƒsd does not eist, and there is no wa to improve the situation b changing ƒ at. The step function in Figure.55d has a jump discontinuit: The one-sided its eist but have different values. The function ƒsd = > in Figure.55e has an infinite discontinuit. The function in Figure.55f has an oscillating discontinuit: It oscillates too much to have a it as :. f() f() f() f() (a) (b) (c) f() sin (d) (e) (f) FIGURE.55 The function in (a) is continuous at = ; the functions in (b) through (f ) are not.

57 .6 Continuit 7 Continuous Functions A function is continuous on an interval if and onl if it is continuous at ever point of the interval. For eample, the semicircle function graphed in Figure.5 is continuous on the interval [-, ], which is its domain. A continuous function is one that is continuous at ever point of its domain. A continuous function need not be continuous on ever interval. For eample, = > is not continuous on [-, ] (Figure.56), but it is continuous over its domain s - q, d s, q d. FIGURE.56 The function = > is continuous at ever value of ecept =. It has a point of discontinuit at = (Eample 5). EXAMPLE 5 Identifing Continuous Functions (a) The function = > (Figure.56) is a continuous function because it is continuous at ever point of its domain. It has a point of discontinuit at =, however, because it is not defined there. (b) The identit function ƒsd = and constant functions are continuous everwhere b Eample 3, Section.3. Algebraic combinations of continuous functions are continuous wherever the are defined. THEOREM 9 Properties of Continuous Functions If the functions ƒ and g are continuous at = c, then the following combinations are continuous at = c.. Sums: ƒ + g. Differences: ƒ - g 3. Products: ƒ # g 4. Constant multiples: k # ƒ, for an number k 5. Quotients: ƒ>g provided gscd Z 6. Powers: f r>s, provided it is defined on an open interval containing c, where r and s are integers Most of the results in Theorem 9 are easil proved from the it rules in Theorem, Section.. For instance, to prove the sum propert we have :c This shows that ƒ + g is continuous. sƒ + gdsd = sƒsd + gsdd :c = ƒsd + gsd, :c :c = ƒscd + gscd = sƒ + gdscd. Sum Rule, Theorem Continuit of ƒ, g at c EXAMPLE 6 Polnomial and Rational Functions Are Continuous (a) Ever polnomial Psd = a n n + a n - n - + Á + a Psd = Pscd b Theorem, Section.. :c is continuous because

58 8 Chapter : Limits and Continuit (b) If P() and Q() are polnomials, then the rational function Psd>Qsd is continuous wherever it is defined sqscd Z d b the Quotient Rule in Theorem 9. EXAMPLE 7 Continuit of the Absolute Value Function The function ƒsd = ƒ ƒ is continuous at ever value of. If 7, we have ƒsd =, a polnomial. If 6, we have ƒsd = -, another polnomial. Finall, at the origin, : ƒ ƒ = = ƒ ƒ. The functions = sin and = cos are continuous at = b Eample 6 of Section.. Both functions are, in fact, continuous everwhere (see Eercise 6). It follows from Theorem 9 that all si trigonometric functions are then continuous wherever the are defined. For eample, = tan is continuous on Á s -p>, p>d sp>, 3p>d Á. Composites All composites of continuous functions are continuous. The idea is that if ƒ() is continuous at = c and g() is continuous at = ƒscd, then g f is continuous at = c (Figure.57). In this case, the it as : c is g(ƒ(c)). g f Continuous at c c f Continuous at c g Continuous at f(c) f(c) g( f(c)) FIGURE.57 Composites of continuous functions are continuous. THEOREM Composite of Continuous Functions If ƒ is continuous at c and g is continuous at ƒ(c), then the composite g f continuous at c. is Intuitivel, Theorem is reasonable because if is close to c, then ƒ() is close to ƒ(c), and since g is continuous at ƒ(c), it follows that g(ƒ()) is close to g(ƒ(c)). The continuit of composites holds for an finite number of functions. The onl requirement is that each function be continuous where it is applied. For an outline of the proof of Theorem, see Eercise 6 in Appendi. EXAMPLE 8 Appling Theorems 9 and Show that the following functions are continuous everwhere on their respective domains. (a) (c) = = ` - - ` (b) = >3 + 4 sin (d) = ` + `

59 .6 Continuit 9 ƒ ƒ FIGURE.58 The graph suggests that = s sin d>s + d is continuous (Eample 8d). Solution (a) The square root function is continuous on [, q d because it is a rational power of the continuous identit function ƒsd = (Part 6, Theorem 9). The given function is then the composite of the polnomial ƒsd = with the square root function gstd = t. (b) The numerator is a rational power of the identit function; the denominator is an everwhere-positive polnomial. Therefore, the quotient is continuous. (c) The quotient s - d>s - d is continuous for all Z ;, and the function is the composition of this quotient with the continuous absolute value function (Eample 7). (d) Because the sine function is everwhere-continuous (Eercise 6), the numerator term sin is the product of continuous functions, and the denominator term + is an everwhere-positive polnomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure.58). Continuous Etension to a Point The function = ssin d> is continuous at ever point ecept =. In this it is like the function = >. But = ssin d> is different from = > in that it has a finite it as : (Theorem 7). It is therefore possible to etend the function s domain to include the point = in such a wa that the etended function is continuous at =. We define The function F() is continuous at = because (Figure.59). Fsd = L sin : sin, Z, =. = Fsd (, ) f() (, ) F(),, (a),, (b) FIGURE.59 The graph (a) of ƒsd = ssin d> for -p> p> does not include the point (, ) because the function is not defined at =. (b) We can remove the discontinuit from the graph b defining the new function F() with Fsd = and Fsd = ƒsd everwhere else. Note that Fsd = ƒsd. :

60 3 Chapter : Limits and Continuit More generall, a function (such as a rational function) ma have a it even at a point where it is not defined. If ƒ(c) is not defined, but :c ƒsd = L eists, we can define a new function F() b the rule Fsd = e ƒsd, if is in the domain of f L, if = c. The function F is continuous at = c. It is called the continuous etension of ƒ to = c. For rational functions ƒ, continuous etensions are usuall found b canceling common factors. EXAMPLE 9 Show that A Continuous Etension (a) (b) FIGURE.6 (a) The graph of ƒ() and (b) the graph of its continuous etension F() (Eample 9). has a continuous etension to =, and find that etension. Solution Although ƒ() is not defined, if Z we have The new function ƒsd = ƒsd = = is equal to ƒ() for Z, but is continuous at =, having there the value of 5>4. Thus F is the continuous etension of ƒ to =, and : - 4 s - ds + 3d s - ds + d = Fsd = = : ƒsd = 5 4. The graph of ƒ is shown in Figure.6. The continuous etension F has the same graph ecept with no hole at (, 5>4). Effectivel, F is the function ƒ with its point of discontinuit at = removed. Intermediate Value Theorem for Continuous Functions Functions that are continuous on intervals have properties that make them particularl useful in mathematics and its applications. One of these is the Intermediate Value Propert. A function is said to have the Intermediate Value Propert if whenever it takes on two values, it also takes on all the values in between. THEOREM The Intermediate Value Theorem for Continuous Functions A function = ƒsd that is continuous on a closed interval [a, b] takes on ever value between ƒ(a) and ƒ(b). In other words, if is an value between ƒ(a) and ƒ(b), then = ƒscd for some c in [a, b].

61 .6 Continuit 3 f(b) f() f(a) a c b FIGURE.6 The function -, 6 ƒsd = e 3, 4 does not take on all values between ƒsd = and ƒs4d = 3; it misses all the values between and 3. Geometricall, the Intermediate Value Theorem sas that an horizontal line = crossing the -ais between the numbers ƒ(a) and ƒ(b) will cross the curve = ƒsd at least once over the interval [a, b]. The proof of the Intermediate Value Theorem depends on the completeness propert of the real number sstem and can be found in more advanced tets. The continuit of ƒ on the interval is essential to Theorem. If ƒ is discontinuous at even one point of the interval, the theorem s conclusion ma fail, as it does for the function graphed in Figure.6. A Consequence for Graphing: Connectivit Theorem is the reason the graph of a function continuous on an interval cannot have an breaks over the interval. It will be connected, a single, unbroken curve, like the graph of sin. It will not have jumps like the graph of the greatest integer function (Figure.54) or separate branches like the graph of > (Figure.56). A Consequence for Root Finding We call a solution of the equation ƒsd = a root of the equation or zero of the function ƒ. The Intermediate Value Theorem tells us that if ƒ is continuous, then an interval on which ƒ changes sign contains a zero of the function. In practical terms, when we see the graph of a continuous function cross the horizontal ais on a computer screen, we know it is not stepping across. There reall is a point where the function s value is zero. This consequence leads to a procedure for estimating the zeros of an continuous function we can graph:. Graph the function over a large interval to see roughl where the zeros are.. Zoom in on each zero to estimate its -coordinate value. You can practice this procedure on our graphing calculator or computer in some of the eercises. Figure.6 shows a tpical sequence of steps in a graphical solution of the equation =.

62 3 Chapter : Limits and Continuit 5.6 (a) (b) (c).3 (d) FIGURE.6 Zooming in on a zero of the function ƒsd = The zero is near =.347.

63 3 Chapter : Limits and Continuit EXERCISES.6 Continuit from Graphs In Eercises 4, sa whether the function graphed is continuous on [-, 3]. If not, where does it fail to be continuous and wh?.. f() g() Eercises 5 are about the function ƒsd = e graphed in the accompaning figure. -, - 6, 6 6, = - + 4, 6 6, f() (, ) 4 (, ) h() k() The graph for Eercises 5.

64 .6 Continuit a. Does ƒs -d eist? b. Does : - + ƒsd eist? c. Does :- + ƒsd = ƒs -d? d. Is ƒ continuous at = -? 6. a. Does ƒ() eist? b. Does : ƒsd eist? c. Does : ƒsd = ƒsd? d. Is ƒ continuous at =? 7. a. Is ƒ defined at =? (Look at the definition of ƒ.) b. Is ƒ continuous at =? 8. At what values of is ƒ continuous? 9. What value should be assigned to ƒ() to make the etended function continuous at =?. To what new value should ƒ() be changed to remove the discontinuit? Appling the Continuit Test At which points do the functions in Eercises and fail to be continuous? At which points, if an, are the discontinuities removable? Not removable? Give reasons for our answers.. Eercise, Section.4. Eercise, Section.4 ƒ ƒ ƒ ƒ At what points are the functions in Eercises 3 8 continuous? 3. = 4. = s + d = 6. = = - + sin 8. = = cos. = cos +. = csc. = tan p = 4 + = tan + + sin 5. = = 3-7. = s - d >3 8. = s - d >5 Composite Functions Find the its in Eercises Are the functions continuous at the point being approached? 9. sin s - sin d :p 3. sin ap cos stan tdb t: 3. sec s : sec - tan - d 3. tan a p : 4 cos ssin >3 db cos a p t: 9-3 sec t b :p>6 csc + 53 tan Continuous Etensions 35. Define g(3) in a wa that etends gsd = s - 9d>s - 3d to be continuous at = Define h() in a wa that etends hstd = st + 3t - d>st - d to be continuous at t =. 37. Define ƒ() in a wa that etends ƒssd = ss 3 - d>ss - d to be continuous at s =. 38. Define g(4) in a wa that etends gsd = s - 6d> s - 3-4d to be continuous at = For what value of a is continuous at ever? 4. For what value of b is continuous at ever? T In Eercises 4 44, graph the function ƒ to see whether it appears to have a continuous etension to the origin. If it does, use Trace and Zoom to find a good candidate for the etended function s value at =. If the function does not appear to have a continuous etension, can it be etended to be continuous at the origin from the right or from the left? If so, what do ou think the etended function s value(s) should be? ƒsd = ƒ ƒ - ƒsd = ƒsd = sin 44. ƒsd = s + d > ƒ ƒ Theor and Eamples ƒsd = e -, 6 3 a, Ú 3, 6 - gsd = e b, Ú A continuous function = ƒsd is known to be negative at = and positive at =. Wh does the equation ƒsd = have at least one solution between = and =? Illustrate with a sketch. 46. Eplain wh the equation cos = has at least one solution. 47. Roots of a cubic Show that the equation = has three solutions in the interval [-4, 4]. 48. A function value Show that the function Fsd = s - ad # s - bd + takes on the value sa + bd> for some value of. 49. Solving an equation If ƒsd = 3-8 +, show that there are values c for which ƒ(c) equals (a) p; (b) - 3; (c) 5,,.

65 34 Chapter : Limits and Continuit 5. Eplain wh the following five statements ask for the same information. a. Find the roots of ƒsd = b. Find the -coordinates of the points where the curve = 3 crosses the line = 3 +. c. Find all the values of for which 3-3 =. d. Find the -coordinates of the points where the cubic curve = 3-3 crosses the line =. e. Solve the equation =. 5. Removable discontinuit Give an eample of a function ƒ() that is continuous for all values of ecept =, where it has a removable discontinuit. Eplain how ou know that ƒ is discontinuous at =, and how ou know the discontinuit is removable. 5. Nonremovable discontinuit Give an eample of a function g() that is continuous for all values of ecept = -, where it has a nonremovable discontinuit. Eplain how ou know that g is discontinuous there and wh the discontinuit is not removable. 53. A function discontinuous at ever point a. Use the fact that ever nonempt interval of real numbers contains both rational and irrational numbers to show that the function ƒsd = e, if is rational, if is irrational is discontinuous at ever point. b. Is ƒ right-continuous or left-continuous at an point? 54. If functions ƒ() and g() are continuous for, could ƒ()>g() possibl be discontinuous at a point of [, ]? Give reasons for our answer. 55. If the product function hsd = ƒsd # gsd is continuous at =, must ƒ() and g() be continuous at =? Give reasons for our answer. 56. Discontinuous composite of continuous functions Give an eample of functions ƒ and g, both continuous at =, for which the composite ƒ g is discontinuous at =. Does this contradict Theorem? Give reasons for our answer. 57. Never-zero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for our answer. T 58. Stretching a rubber band Is it true that if ou stretch a rubber band b moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for our answer. 59. A fied point theorem Suppose that a function ƒ is continuous on the closed interval [, ] and that ƒsd for ever in [, ]. Show that there must eist a number c in [, ] such that ƒscd = c (c is called a fied point of ƒ). 6. The sign-preserving propert of continuous functions Let ƒ be defined on an interval (a, b) and suppose that ƒscd Z at some c where ƒ is continuous. Show that there is an interval sc - d, c + dd about c where ƒ has the same sign as ƒ(c). Notice how remarkable this conclusion is. Although ƒ is defined throughout (a, b), it is not required to be continuous at an point ecept c. That and the condition ƒscd Z are enough to make ƒ different from zero (positive or negative) throughout an entire interval. 6. Prove that ƒ is continuous at c if and onl if ƒsc + hd = ƒscd. h: 6. Use Eercise 6 together with the identities sin sh + cd = sin h cos c + cos h sin c, cos sh + cd = cos h cos c - sin h sin c to prove that ƒsd = sin and gsd = cos are continuous at ever point = c. Solving Equations Graphicall Use a graphing calculator or computer grapher to solve the equations in Eercises = = 65. s - d = sone rootd 66. = = = sthree rootsd 69. cos = sone rootd. Make sure ou are using radian mode. 7. sin = sthree rootsd. Make sure ou are using radian mode.

66 34 Chapter : Limits and Continuit.7 Tangents and Derivatives This section continues the discussion of secants and tangents begun in Section.. We calculate its of secant slopes to find tangents to curves. What Is a Tangent to a Curve? For circles, tangenc is straightforward. A line L is tangent to a circle at a point P if L passes through P perpendicular to the radius at P (Figure.63). Such a line just touches

67 .7 Tangents and Derivatives 35 O P FIGURE.63 L is tangent to the circle at P if it passes through P perpendicular to radius OP. L the circle. But what does it mean to sa that a line L is tangent to some other curve C at a point P? Generalizing from the geometr of the circle, we might sa that it means one of the following:. L passes through P perpendicular to the line from P to the center of C.. L passes through onl one point of C, namel P. 3. L passes through P and lies on one side of C onl. Although these statements are valid if C is a circle, none of them works consistentl for more general curves. Most curves do not have centers, and a line we ma want to call tangent ma intersect C at other points or cross C at the point of tangenc (Figure.64). L P C C L L P C P L meets C onl at P but is not tangent to C. L is tangent to C at P but meets C at several points. L is tangent to C at P but lies on two sides of C, crossing C at P. FIGURE.64 Eploding mths about tangent lines. HISTORICAL BIOGRAPHY Pierre de Fermat (6 665) To define tangenc for general curves, we need a dnamic approach that takes into account the behavior of the secants through P and nearb points Q as Q moves toward P along the curve (Figure.65). It goes like this:. We start with what we can calculate, namel the slope of the secant PQ.. Investigate the it of the secant slope as Q approaches P along the curve. 3. If the it eists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope. This approach is what we were doing in the falling-rock and fruit fl eamples in Section.. Tangent Secants P P Q Tangent Secants Q FIGURE.65 The dnamic approach to tangenc. The tangent to the curve at P is the line through P whose slope is the it of the secant slopes as Q : P from either side.

68 36 Chapter : Limits and Continuit EXAMPLE Tangent Line to a Parabola Find the slope of the parabola = at the point P(, 4). Write an equation for the tangent to the parabola at this point. Solution We begin with a secant line through P(, 4) and Qs + h, s + hd d nearb. We then write an epression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve: Secant slope = = s + hd - h If h 7, then Q lies above and to the right of P, as in Figure.66. If h 6, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope approaches 4: We take 4 to be the parabola s slope at P. The tangent to the parabola at P is the line through P with slope 4: = 4 + 4s - d = 4-4. sh + 4d = 4. h: = h + 4h h = h + 4h h Point-slope equation = h + 4. Q( h, ( h) ) ( h) Secant slope is 4 h 4. h Tangent slope 4 ( h) 4 P(, 4) h h NOT TO SCALE FIGURE.66 Finding the slope of the parabola = at the point P(, 4) (Eample ). Finding a Tangent to the Graph of a Function The problem of finding a tangent to a curve was the dominant mathematical problem of the earl seventeenth centur. In optics, the tangent determined the angle at which a ra of light entered a curved lens. In mechanics, the tangent determined the direction of a bod s motion at ever point along its path. In geometr, the tangents to two curves at a point of intersection determined the angles at which the intersected. To find a tangent to an arbitrar curve = ƒsd at a point Ps, ƒs dd, we use the same dnamic procedure. We calculate the slope of the secant through P and a point Qs + h, ƒs + hdd. We then investigate the it of the slope as h : (Figure.67). If the it eists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.

69 .7 Tangents and Derivatives 37 f() Q( h, f( h)) f( h) f( ) P(, f( )) DEFINITIONS Slope, Tangent Line The slope of the curve = ƒsd at the point Ps, ƒsdd is the number ƒs + hd - ƒs d m = (provided the it eists). h: h The tangent line to the curve at P is the line through P with this slope. h h FIGURE.67 The slope of the tangent ƒs + hd - ƒs d line at P is. h: h Whenever we make a new definition, we tr it on familiar objects to be sure it is consistent with results we epect in familiar cases. Eample shows that the new definition of slope agrees with the old definition from Section. when we appl it to nonvertical lines. EXAMPLE Testing the Definition Show that the line = m + b is its own tangent at an point s, m + bd. Solution We let ƒsd = m + b and organize the work into three steps.. Find ƒs d and ƒs + hd. ƒs d = m + b. Find the slope sƒs + hd - ƒs dd>h. h: ƒs + hd - ƒs d sm + mh + bd - sm + bd = h: h h: h 3. Find the tangent line using the point-slope equation. The tangent line at the point s, m + bd is = m + b. Let s summarize the steps in Eample. ƒs + hd = ms + hd + b = m + mh + b mh = h: h = m = sm + bd + ms - d = m + b + m - m Finding the Tangent to the Curve ƒsd at s, d. Calculate ƒs d and ƒs + hd.. Calculate the slope ƒs + hd - ƒs d m =. h: h 3. If the it eists, find the tangent line as = + ms - d.

70 38 Chapter : Limits and Continuit EXAMPLE 3 Slope and Tangent to = >, Z (a) Find the slope of the curve = > at = a Z. (b) Where does the slope equal ->4? (c) What happens to the tangent to the curve at the point (a, >a) as a changes? Solution (a) Here ƒsd = >. The slope at (a, >a) is ƒsa + hd - ƒsad h: h = h: a - sa + hd = h: h asa + hd = h: -h hasa + hd = h: - asa + hd =- a. Notice how we had to keep writing h: before each fraction until the stage where we could evaluate the it b substituting h =. The number a ma be positive or negative, but not. (b) The slope of = > at the point where = a is ->a. It will be ->4 provided that - a =- 4. a + h - a This equation is equivalent to a = 4, so a = or a = -. The curve has slope ->4 at the two points (, >) and s -, ->d (Figure.68). (c) Notice that the slope ->a is alwas negative if a Z. As a : +, the slope approaches - q and the tangent becomes increasingl steep (Figure.69). We see this situation again as a : -. As a moves awa from the origin in either direction, the slope approaches - and the tangent levels off. h slope is a slope is 4, a, slope is 4 FIGURE.68 The two tangent lines to = > having slope ->4 (Eample 3). FIGURE.69 The tangent slopes, steep near the origin, become more gradual as the point of tangenc moves awa.

71 .7 Tangents and Derivatives 39 Rates of Change: Derivative at a Point The epression ƒs + hd - ƒs d h is called the difference quotient of ƒ at with increment h. If the difference quotient has a it as h approaches zero, that it is called the derivative of ƒ at. If we interpret the difference quotient as a secant slope, the derivative gives the slope of the curve and tangent at the point where =. If we interpret the difference quotient as an average rate of change, as we did in Section., the derivative gives the function s rate of change with respect to at the point =. The derivative is one of the two most important mathematical objects considered in calculus. We begin a thorough stud of it in Chapter 3. The other important object is the integral, and we initiate its stud in Chapter 5. EXAMPLE 4 Instantaneous Speed (Continuation of Section., Eamples and ) In Eamples and in Section., we studied the speed of a rock falling freel from rest near the surface of the earth. We knew that the rock fell = 6t feet during the first t sec, and we used a sequence of average rates over increasingl short intervals to estimate the rock s speed at the instant t =. Eactl what was the rock s speed at this time? Solution We let ƒstd = 6t. The average speed of the rock over the interval between t = and t = + h seconds was ƒs + hd - ƒsd h = 6s + hd - 6sd h The rock s speed at the instant t = was 6sh + d = 6s + d = 3 ft>sec. h: Our original estimate of 3 ft > sec was right. = 6sh + hd h = 6sh + d. Summar We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, the it of the difference quotient, and the derivative of a function at a point. All of these ideas refer to the same thing, summarized here:. The slope of = ƒsd at =. The slope of the tangent to the curve = ƒsd at = 3. The rate of change of ƒ() with respect to at = 4. The derivative of ƒ at = 5. ƒs + hd - ƒs d The it of the difference quotient, h: h

72 4 Chapter : Limits and Continuit EXERCISES.7 Slopes and Tangent Lines In Eercises 4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in -units per -unit) at the points P and P. Graphs can shift during a press run, so our estimates ma be somewhat different from those in the back of the book P P P In Eercises 5, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. 5. = 4 -, s -, 3d 6. = s - d +, s, d 7. =, s, d 8. P =, s -, d 9. =. = 3, a-, - 3, s -, -8d 8 b P In Eercises 8, find the slope of the function s graph at the given point. Then find an equation for the line tangent to the graph there.. ƒsd = +, s, 5d. ƒsd = -, s, -d 3. gsd =, s3, 3d - 4. gsd = 8, s, d 5. hstd = t 3, s, 8d 6. hstd = t 3 + 3t, s, 4d 7. ƒsd =, s4, d 8. ƒsd = +, s8, 3d 4 3 P P P In Eercises 9, find the slope of the curve at the point indicated. 9. = 5, = -. = -, =. = -, = 3. = - +, = Tangent Lines with Specified Slopes At what points do the graphs of the functions in Eercises 3 and 4 have horizontal tangents? 3. ƒsd = gsd = Find equations of all lines having slope - that are tangent to the curve = >s - d. 6. Find an equation of the straight line having slope >4 that is tangent to the curve =. Rates of Change 7. Object dropped from a tower An object is dropped from the top of a -m-high tower. Its height above ground after t sec is - 4.9t m. How fast is it falling sec after it is dropped? 8. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t ft. How fast is the rocket cbing sec after liftoff? 9. Circle s changing area What is the rate of change of the area of a circle sa = pr d with respect to the radius when the radius is r = 3? 3. Ball s changing volume What is the rate of change of the volume of a ball sv = s4>3dpr 3 d with respect to the radius when the radius is r =? Testing for Tangents 3. Does the graph of have a tangent at the origin? Give reasons for our answer. 3. Does the graph of have a tangent at the origin? Give reasons for our answer. Vertical Tangents ƒsd = e sin s>d, Z, = sin s>d, Z gsd = e, = We sa that the curve = ƒsd has a vertical tangent at the point where = if h: sƒs + hd - ƒs dd>h = q or - q. Vertical tangent at = (see accompaning figure): ƒs + hd - ƒsd h = >3 - h: h h: h = h: h = q >3

73 .7 Tangents and Derivatives 4 No vertical tangent at = (see net figure): gs + hd - gsd h = >3 - h: h h: h does not eist, because the it is q from the right and - q from the left. 33. Does the graph of VERTICAL TANGENT AT ORIGIN = h: h >3 ƒsd = f() 3 g() 3 NO VERTICAL TANGENT AT ORIGIN -, 6, =, 7 have a vertical tangent at the origin? Give reasons for our answer. 34. Does the graph of ƒ ƒ ƒ have a vertical tangent at the point (, )? Give reasons for our answer. T a. Graph the curves in Eercises Where do the graphs appear to have vertical tangents? b. Confirm our findings in part (a) with it calculations. But before ou do, read the introduction to Eercises 33 and = >5 36. = 4>5 37. = >5 38. = 3>5 39. = 4 >5-4. = 5>3-5 >3 4. = >3 - s - d >3 4. = >3 + s - d >3 43. = e - ƒ,, = 4 - COMPUTER EXPLORATIONS Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Eercises a. Plot = ƒsd over the interval s - >d s + 3d. b. Holding fied, the difference quotient qshd = ƒs + hd - ƒs d h Usd = e, 6, Ú at becomes a function of the step size h. Enter this function into our CAS workspace. c. Find the it of q as h :. d. Define the secant lines = ƒs d + q # s - d for h = 3,, and. Graph them together with ƒ and the tangent line over the interval in part (a). 45. ƒsd = 3 +, 46. ƒsd = + 5 =, = 47. ƒsd = + sin sd, = p> 48. ƒsd = cos + 4 sin sd, = p

74 44 Chapter : Limits and Continuit Chapter Additional and Advanced Eercises T T. Assigning a value to The rules of eponents (see Appendi 9) tell us that a = if a is an number different from zero. The also tell us that n = if n is an positive number. If we tried to etend these rules to include the case, we would get conflicting results. The first rule would sa =, whereas the second would sa =. We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, define to have an value we wanted as long as we could persuade others to agree. What value would ou like to have? Here is an eample that might help ou to decide. (See Eercise below for another eample.) a. Calculate for =.,.,., and so on as far as our calculator can go. Record the values ou get. What pattern do ou see? b. Graph the function = for 6. Even though the function is not defined for, the graph will approach the -ais from the right. Toward what -value does it seem to be headed? Zoom in to further support our idea.. A reason ou might want to be something other than or As the number increases through positive values, the numbers > and > (ln ) both approach zero. What happens to the number ƒsd = a b >sln d as increases? Here are two was to find out. a. Evaluate ƒ for =,,, and so on as far as our calculator can reasonabl go. What pattern do ou see? b. Graph ƒ in a variet of graphing windows, including windows that contain the origin. What do ou see? Trace the -values along the graph. What do ou find? 3. Lorentz contraction In relativit theor, the length of an object, sa a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket s length as L at rest, then at speed the length will appear to be L = L B - c. Eit rate ft 3 min Suppose that = > for a certain tank. You are tring to maintain a fairl constant eit rate b adding water to the tank with a hose from time to time. How deep must ou keep the water if ou want to maintain the eit rate a. within. ft 3 >min of the rate = ft 3 >min? b. within. ft 3 >min of the rate = ft 3 >min? 5. Thermal epansion in precise equipment As ou ma know, most metals epand when heated and contract when cooled. The dimensions of a piece of laborator equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laborator where the equipment is to be used. A tpical aluminum bar that is cm wide at 7 F will be = + st - 7d * -4 centimeters wide at a nearb temperature t. Suppose that ou are using a bar like this in a gravit wave detector, where its width must sta within.5 cm of the ideal cm. How close to t = 7 F must ou maintain the temperature to ensure that this tolerance is not eceeded? 6. Stripes on a measuring cup The interior of a tpical -L measuring cup is a right circular clinder of radius 6 cm (see accompaning figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being V = p6 h = 36ph. How closel must we measure h to measure out L of water s cm 3 d with an error of no more than % s cm 3 d? This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 * 8 m>sec. What happens to L as increases? Find :c - L. Wh was the left-hand it needed? 4. Controlling the flow from a draining tank Torricelli s law sas that if ou drain a tank like the one in the figure shown, the rate at which water runs out is a constant times the square root of the water s depth. The constant depends on the size and shape of the eit valve. Stripes about mm wide (a)

75 Chapter Additional and Advanced Eercises 45 h (b) r 6 cm Liquid volume V 36h A -L measuring cup (a), modeled as a right circular clinder (b) of radius r = 6 cm Precise Definition of Limit ƒ ƒ ƒ In Eercises 7, use the formal definition of it to prove that the function is continuous at. 7. ƒsd = - 7, = 8. gsd = >sd, = >4 9. hsd = - 3, =. Fsd = 9 -, = 5. Uniqueness of its Show that a function cannot have two different its at the same point. That is, if : ƒsd = L and : ƒsd = L, then L = L.. Prove the it Constant Multiple Rule: k ƒsd for an constant k. :c :c 3. One-sided its If : + ƒsd = A and : - ƒsd = B, find a. : + ƒs 3 - d b. : - ƒs 3 - d c. : + ƒs - 4 d d. : - ƒs - 4 d 4. Limits and continuit Which of the following statements are true, and which are false? If true, sa wh; if false, give a countereample (that is, an eample confirming the falsehood). a. If :a ƒsd eists but :a gsd does not eist, then :a sƒsd + gsdd does not eist. b. If neither :a ƒsd nor :a gsd eists, then :a sƒsd + gsdd does not eist. c. If ƒ is continuous at, then so is ƒ ƒ. d. If ƒ is continuous at a, then so is ƒ. In Eercises 5 and 6, use the formal definition of it to prove that the function has a continuous etension to the given value of gsd = ƒsd = -, = 3 +, = A function continuous at onl one point Let ƒsd = e, if is rational, if is irrational. a. Show that ƒ is continuous at =. b. Use the fact that ever nonempt open interval of real numbers contains both rational and irrational numbers to show that ƒ is not continuous at an nonzero value of. 8. The Dirichlet ruler function If is a rational number, then can be written in a unique wa as a quotient of integers m>n where n 7 and m and n have no common factors greater than. (We sa that such a fraction is in lowest terms. For eample, 6>4 written in lowest terms is 3>.) Let ƒ() be defined for all in the interval [, ] b ƒsd = e >n, if = m>n is a rational number in lowest terms, if is irrational. For instance, ƒsd = ƒsd =, ƒs>d = >, ƒs>3d = ƒs>3d = >3, ƒs>4d = ƒs3>4d = >4, and so on. a. Show that ƒ is discontinuous at ever rational number in [, ]. b. Show that ƒ is continuous at ever irrational number in [, ]. (Hint: If P is a given positive number, show that there are onl finitel man rational numbers r in [, ] such that ƒsrd ÚP. ) c. Sketch the graph of ƒ. Wh do ou think ƒ is called the ruler function? 9. Antipodal points Is there an reason to believe that there is alwas a pair of antipodal (diametricall opposite) points on Earth s equator where the temperatures are the same? Eplain.. If sƒsd + gsdd = 3 and sƒsd - gsdd = -, find :c ƒsdgsd. :c. Roots of a quadratic equation that is almost linear The equation a + - =, where a is a constant, has two roots if a 7 - and a Z, one positive and one negative: r + sad = :c a a, r - sad = a a. a. What happens to r as As a : sad a :?? b. What happens to r as As a : sad a :?? c. Support our conclusions b graphing r + sad and r - sad as functions of a. Describe what ou see. d. For added support, graph ƒsd = a + - simultaneousl for a =,.5,.,., and.5.. Root of an equation Show that the equation + cos = has at least one solution. 3. Bounded functions A real-valued function ƒ is bounded from above on a set D if there eists a number N such that ƒsd N for all in D. We call N, when it eists, an upper bound for ƒ on D and sa that ƒ is bounded from above b N. In a similar manner, we sa that ƒ is bounded from below on D if there eists a number M such that ƒsd Ú M for all in D. We call M, when it eists, a lower bound for ƒ on D and sa that ƒ is bounded from below b M. We sa that ƒ is bounded on D if it is bounded from both above and below. a. Show that ƒ is bounded on D if and onl if there eists a number B such that ƒ ƒsd ƒ B for all in D. b. Suppose that ƒ is bounded from above b N. Show that if : ƒsd = L, then L N. c. Suppose that ƒ is bounded from below b M. Show that if : ƒsd = L, then L Ú M.