or Use Onl in Pilot Program F 96 Chapter Limits 6 7. Steep secant lines a. Given the graph of f in the following figures, find the slope of the secant line that passes through, and h, f h in terms of h, for h 7 and h 6.. Evaluate the it of the slope of the secant line found in part (a) as h S + and h S -. What does this tell ou aout the line tangent to the curve at,?. Create a graph that gives a more complete representation of f. 6. f = > 7. f = > / (, ) / (, ) h h (h, h / ) (h, h / ) T 8. Care with graphing The figure shows the graph of the function f = graphed in the window -, *,. + a. Evaluate f, f, and f. S + S - S T Technolog Eercises 9 6. Asmptotes Use analtical methods and/or a graphing utilit to identif the vertical asmptotes (if an) of the following functions. 9. f = - + - 9. g = - ln e. h = +. p = sec ap, for 6. gu = tan a pu. qs = p s - sin s. f = 6. g = e > sec T 7. Can a graph intersect a vertical asmptote? A common misconception is that the graph of a function never intersects its vertical asmptotes. Let f = - if 6 if Ú. Eplain wh = is a vertical asmptote of the graph of f and show that the graph of f intersects the line =. QUICK CHECK ANSWERS. Answers will var, ut all graphs should have a vertical asmptote at =.. - ;. As S - +, 6 and + 7, so + S through negative values. - -. = - =, which is not S - S an infinite it, so = is not a vertical asmptote.. Limits at Infinit Limits at infinit as opposed to infinite its occur when the independent variale ecomes large in magnitude. For this reason, its at infinit determine what is called the end ehavior of a function. An application of these its is to determine whether a sstem (such as an ecosstem or a large oscillating structure) reaches a stead state as time increases. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F. Limits at Infinit 97 Horizontal asmptote q f () q f () tan Limits at Infinit and Horizontal Asmptotes Consider the function f = tan -, whose domain is -, (Figure.). As ecomes aritraril large (denoted S ), f approaches p>, and as ecomes aritraril large in magnitude and negative (denoted S - ), f approaches -p>. These its are epressed as q f () q FIGURE. f () M FIGURE. L f () f () M Horizontal asmptote f () L f () S tan- = p and S- tan- = - p. The graph of f approaches the horizontal line = p> as S, and it approaches the horizontal line = -p> as S -. These lines are called horizontal asmptotes. DEFINITION Limits at Infinit and Horizontal Asmptotes If f ecomes aritraril close to a finite numer L for all sufficientl large and positive, then we write S f = L. We sa the it of f as approaches infinit is L. In this case, the line = L is a horizontal asmptote of f (Figure.). The it at negative infinit, S - f = M, is defined analogousl. When the it eists, the horizontal asmptote is = M. QUICK CHECK Evaluate > + for =,, and. What is S +? The it laws of Theorem. and the Squeeze Theorem appl if S a is replaced with S or S -. f () f () f () FIGURE. EXAMPLE Limits at infinit Evaluate the following its. a. a + S - SOLUTION sin. a + S a. As ecomes large and negative, ecomes large and positive; in turn, > approaches. B the it laws of Theorem., a + S- = + a S- S- = + =. (++)++* equals equals c Notice that a + is also equal to. Therefore, the graph of = + > S approaches the horizontal asmptote = as S and as S - (Figure.).. The numerator of sin > is ounded etween - and ; therefore, for 7, - sin. As S, ecomes aritraril large, which means that S - = S =. sin It follows the Squeeze Theorem (Theorem.) that S =. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F 98 Chapter Limits 6 sin f () f () Using the it laws of Theorem., sin a + S = + asin S S =. e(++)++* equals equals The graph of = + sin approaches the horizontal asmptote = as ecomes large (Figure.). Note that the curve intersects its asmptote infinitel man times. Related Eercises 9 FIGURE. f () Infinite Limits at Infinit It is possile for a it to e oth an infinite it and a it at infinit. This tpe of it occurs if f ecomes aritraril large in magnitude as ecomes aritraril large in magnitude. Such a it is called an infinite it at infinit and is illustrated the function f = (Figure.). f () DEFINITION Infinite Limits at Infinit If f ecomes aritraril large as ecomes aritraril large, then we write f =. S The its f = -, f =, and f = - are S S - S - defined similarl. f () FIGURE. Infinite its at infinit tell us aout the ehavior of polnomials for large-magnitude values of. First, consider power functions f = n, where n is a positive integer. Figure. shows that when n is even, S { n =, and when n is odd, n = and S S - n = -. n even: n 6 6 n odd: n n 7 FIGURE. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F. Limits at Infinit 99 QUICK CHECK Descrie the ehavior of p = - as S and as S -. It follows that reciprocals of power functions f = > n = -n, where n is a positive integer, ehave as follows: S n = S -n = and S - n = S - -n =. From here, it is a short step to finding the ehavior of an polnomial as S {. Let p = a n n + a n - n - + g + a + a + a. We now write p in the equivalent form p = n a n + a n - e S + a n - + g+ a n. " e S S Notice that as ecomes large in magnitude, all the terms in p ecept the first term approach zero. Therefore, as S {, we see that p a n n. This means that as S {, the ehavior of p is determined the term a n n with the highest power of. THEOREM.6 Limits at Infinit of Powers and Polnomials Let n e a positive integer and let p e the polnomial p = a n n + a n - n - + g + a + a + a, where a n.. n = when n is even. S {. n = and n = - when n is odd. S S -. S { n = S { -n =.. p = a n n = or -, depending on the degree of the S { S { polnomial and the sign of the leading coefficient a n. EXAMPLE Limits at infinit Evaluate the its as S { of the following functions. a. p = - 6 + -. q = - + - SOLUTION a. We use the fact that the it is determined the ehavior of the leading term: - 6 + - = =. S S S Similarl, S - - 6 + - = S - =. S. Noting that the leading coefficient is negative, we have S - + - = - = - S S - + - = - S - S - =. S - Related Eercises Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits End Behavior The ehavior of polnomials as S { is an eample of what is often called end ehavior. Having treated polnomials, we now turn to the end ehavior of rational, algeraic, and transcendental functions. EXAMPLE End ehavior of rational functions Determine the end ehavior for the following rational functions. a. f = + - SOLUTION. g = + - + 8 + c. h = - + + a. An effective approach for evaluating its of rational functions at infinit is to divide oth the numerator and denominator n, where n is the largest power appearing in the denominator. This strateg forces the terms corresponding to lower powers of to approach in the it. In this case, we divide : approaches Recall that the degree of a polnomial is the highest power of that appears. f () FIGURE.6 g() f () FIGURE.7 f () g() g() 8 + + S - = S - + = S - = approaches =. + A similar calculation gives S - =, and thus the graph of f has the horizontal - asmptote =. You should confirm that the zeros of the denominator are - and, which correspond to vertical asmptotes (Figure.6). In this eample, the degree of the polnomial in the numerator is less than the degree of the polnomial in the denominator.. Again we divide oth the numerator and denominator the largest power appearing in the denominator, which is : S + - + 8 + = S + - Divide the numerator and + 8 + denominator. approaches approaches + - = S + 8 + Simplif. = + + + + approaches approaches =. Evaluate its. Using the same steps (dividing each term ), it can e shown that + - S - + 8 =. This function has the horizontal asmptote = + (Figure.7). Notice that the degree of the polnomial in the numerator equals the degree of the polnomial in the denominator. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F. Limits at Infinit c. We divide the numerator and denominator the largest power of appearing in the denominator, which is, and then take the it: - + S + = S - + + aritraril large constant approaches Divide numerator and denominator. - + = S + Simplif. constant approaches =. Take its. As S, all the terms in this function either approach zero or are constant ecept the @term in the numerator, which ecomes aritraril large. Therefore, the it of the - + function does not eist. Using a similar analsis, we find that =. S- + These its are not finite, and so the graph of the function has no horizontal asmptote. In this case, the degree of the polnomial in the numerator is greater than the degree of the polnomial in the denominator. Related Eercises The conclusions reached in Eample can e generalized for all rational functions. These results are summarized in Theorem.7 (Eercise 7). QUICK CHECK Use Theorem.7 to find the vertical and horizontal asmptotes of = -. THEOREM.7 End Behavior and Asmptotes of Rational Functions Suppose f = p is a rational function, where q p = a m m + a m - m - + g + a + a + a and q = n n + n - n - + g + + +, with a m and n. a. Degree of numerator less than degree of denominator If m 6 n, then S { f =, and = is a horizontal asmptote of f.. Degree of numerator equals degree of denominator If m = n, then S { f = a m > n, and = a m > n is a horizontal asmptote of f. c. Degree of numerator greater than degree of denominator If m 7 n, then S { f = or -, and f has no horizontal asmptote. d. Assuming that f is in reduced form (p and q share no common factors), vertical asmptotes occur at the zeros of q. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits Although it isn t stated eplicitl, Theorem.7 implies that a rational function can have at most one horizontal asmptote, and whenever there is a horizontal asmptote, p S q = p. The same cannot e said of other functions, as the net eamples S - q show. Recall that = = if Ú - if 6. Therefore, 6 = = if Ú - if 6. Because is negative as S -, we have 6 = -. f () 8 6 FIGURE.8 6 f () f () EXAMPLE End ehavior of an algeraic function Eamine the end ehavior of f = - + 8 6 + +. SOLUTION The square root in the denominator forces us to revise the strateg used with rational functions. First, consider the it as S. The highest power of the polnomial in the denominator is 6. However, the polnomial is under a square root, so effectivel, the highest power in the denominator is 6 =. Dividing the numerator and denominator, for 7, the it is evaluated as follows: S - + 8 6 + + = S = S - + 8 6 B 6 + 6 + Divide 6 =. 6 approaches approaches - 8 + Simplif. A + + 6 approaches approaches = =. Evaluate its. As S -, is negative, so we divide numerator and denominator 6 = - (which is positive): S - - + 8 6 + + = S - - - - + 8-6 B 6 + 6 + 6 Divide 6 = - 7. approaches approaches - + 8 - = S - A + + 6 Simplif. approaches approaches = - = -. Evaluate its. The its reveal two asmptotes, = and = -. Oserve that the graph crosses oth horizontal asmptotes (Figure.8). Related Eercises 8 EXAMPLE End ehavior of transcendental functions Determine the end ehavior of the following transcendental functions. a. f = e and g = e -. h = ln c. f = cos Copright Pearson Education, Inc.
or Use Onl in Pilot Program F. Limits at Infinit e FIGURE.9 FIGURE. f () e f () e e ln h() f () ln Reflection of e across line ln Tale.9 SOLUTION a. The graph of f = e (Figure.9) makes it clear that as S, e increases without ound. All eponential functions with 7 ehave this wa, ecause raising a numer greater than to ever-larger powers produces numers that increase without ound. The figure also suggests that as S -, the graph of e approaches the horizontal asmptote =. This claim is confirmed analticall recognizing that S - e = e - = S S e =. Therefore, e = and S S - e =. Because e - = >e, it follows that S e- = and S - e- =.. The domain of ln is : 7 6, so we evaluate S +ln and ln to determine S end ehavior. For the first it, recall that ln is the inverse of e (Figure.), and the graph of ln is a reflection of the graph of e across the line =. The horizontal asmptote = of e is also reflected across =, ecoming a vertical asmptote = for ln. These oservations impl that ln = -. + S It is not ovious whether the graph of ln approaches a horizontal asmptote or whether the function grows without ound as S. Furthermore, the numerical evidence (Tale.9) is inconclusive ecause ln increases ver slowl. The inverse relation etween e and ln is again useful. The fact that the domain of e is -, implies that the range of ln is also -,. Therefore, the values of ln lie in the interval -,, and it follows that ln =. S c. The cosine function oscillates etween - and as approaches infinit (Figure.). Therefore, cos does not eist. For the same reason, cos does not eist. S S - ln...6.9 99 7.96 T T??? cos does not eist. f () cos cos does not eist. FIGURE. Related Eercises 9 The end ehavior of eponential and logarithmic functions are important in upcoming work. We summarize these results in the following theorem. QUICK CHECK How do the functions e and e - ehave as S and as S -? THEOREM.8 End Behavior of e, e -, and ln The end ehavior for e and e - on -, and ln on, is given the following its: S e = and S - e =, S e- = and S - e- =, ln = - and ln =. + S S Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits SECTION. EXERCISES Review Questions. Eplain the meaning of f =. S -. What is a horizontal asmptote? f. Determine S g if f S, and g S as S.. Descrie the end ehavior of g = e -.. Descrie the end ehavior of f = -. 6. The tet descries three cases that arise when eamining the end ehavior of a rational function f = p>q. Descrie the end ehavior associated with each case. 7. Evaluate e, e, and e -. S S - S 8. Use a sketch to find the end ehavior of f = ln. Basic Skills 9. Limits at infinit Evaluate the following its. 9. S a +. a + S + cos u + +.. us u S cos. S. a + S - + sin. Infinite its at infinit Determine the following its.. 6. S S- 7. - 6 8. - S S- 9. - 9 7. 7 + S S -. - 6 +. -8 S - S -. - -. -8 + S S -. Rational functions Evaluate f and f for the S S - following rational functions. Then give the horizontal asmptote of f (if an).. f = + 7. f = 6-9 + 8 + 9. f = - 7 +. f =. f = + -. f = + 6-6. f = - 7 + - 7 8. f = 8 + + + 7 + -. f = 8-8 - 7. f = - + + 8 8. Algeraic functions Evaluate f and f for the S S - following functions. Then give the horizontal asmptote(s) of f (if an). +. f = + 6 6 + 6. f = + + 6 + 8 7. f = + + 8. f = - 9 + 9. Transcendental functions Determine the end ehavior of the following transcendental functions evaluating appropriate its. Then provide a simple sketch of the associated graph, showing asmptotes if the eist. 9. f = -e -. f =. f = - ln. f = ln. f = sin. f = e Further Eplorations. Eplain wh or wh not Determine whether the following statements are true and give an eplanation or countereample. a. The graph of a function can never cross one of its horizontal asmptotes.. A rational function f can have oth f = L and S f =. S - c. The graph of an function can have at most two horizontal asmptotes. 6. Horizontal and vertical asmptotes a. Evaluate f and S S - f, and then identif an horizontal asmptotes.. Find the vertical asmptotes. For each vertical asmptote = a, evaluate Sa - f and Sa + f. 6. f = - + - 8. f = 6 + 6 + - 9. f = + - 6 - +. f = 6-6 +. f = - 9 -. f = - > -. f = + + 6 - - 7. f = + + + Copright Pearson Education, Inc.
or Use Onl in Pilot Program F. Limits at Infinit T. f = - +. f = - - 6 9. End ehavior for transcendental functions 6. The central ranch of f = tan is shown in the figure. a. Evaluate Sp>-tan and S-p> +tan. Are these infinite its or its at infinit?. Sketch a graph of g = tan - reflecting the graph of f over the line =, and use it to evaluate tan - and S tan -. S - q f () tan 7. Graph = sec - and evaluate the following its using the graph. Assume the domain is : Ú 6. a. sec -. sec - S S - 8. The hperolic cosine function, denoted cosh, is used to model the shape of a hanging cale (a telephone wire, for eample). It is defined as cosh = e + e -. a. Determine its end ehavior evaluating cosh and S cosh. S -. Evaluate cosh. Use smmetr and part (a) to sketch a plausile graph for = cosh. 9. The hperolic sine function is defined as sinh = e - e -. a. Determine its end ehavior evaluating sinh and S sinh. S -. Evaluate sinh. Use smmetr and part (a) to sketch a plausile graph for = sinh. 6 6. Sketching graphs Sketch a possile graph of a function f that satisfies all the given conditions. Be sure to identif all vertical and horizontal asmptotes. 6. f - = -, f =, f =, f =, S f = - S - q 6. f =, f = -, f =, S + S - S f = - S - 6. Asmptotes Find the vertical and horizontal asmptotes of f = e >. 6. Asmptotes Find the vertical and horizontal asmptotes of cos + f =. Applications 6 69. Stead states If a function f represents a sstem that varies in time, the eistence of f t means that the sstem reaches a stead ts state (or equilirium). For the following sstems, determine if a stead state eists and give the stead-state value. 6. The population of a acteria culture is given pt = t +. 6. The population of a culture of tumor cells is given pt = t t +. 66. The amount of drug (in milligrams) in the lood after an IV tue is inserted is mt = - -t. 67. The value of an investment in dollars is given vt = e.6t. 68. The population of a colon of squirrels is given pt = + e -.t. 69. The amplitude of an oscillator is given at = a t + sin t. t 7 7. Looking ahead to sequences A sequence is an infinite, ordered list of numers that is often defined a function. For eample, the sequence,, 6, 8, c6 is specified the function f n = n, where n =,,, c. The it of such a sequence is ns f n, provided the it eists. All the it laws for its at infinit ma e applied to its of sequences. Find the it of the following sequences, or state that the it does not eist. 7. e,,,,,, cf, which is defined f n = n, for n =,,, c 7. e,,,, cf, which is defined f n = n -, n for n =,,, c 7. e,, 9, 6, cf, which is defined f n = n n +, for n =,,, c 7. e,, 9, 6, cf, which is defined f n = n + n, for n =,,, c Copright Pearson Education, Inc.
or Use Onl in Pilot Program F 6 Chapter Limits T Additional Eercises 7. End ehavior of a rational function Suppose f = p q is a rational function, where p = a m m + a m - m - + g + a + a + a, q = n n + n - n - + g + + +, a m, and n. a. Prove that if m = n, then f = a m. S { n. Prove that if m 6 n, then f =. S { 7 76. Limits of eponentials Evaluate f and f. S S- Then state the horizontal asmptote(s) of f. Confirm our findings plotting f. 7. f = e + e e + e 76. f = e + e - e + e - T 77. Sutle asmptotes Use analtical methods to identif all the asmptotes of f = ln 9 - e -. Then confirm our results - e locating the asmptotes using a graphing calculator. QUICK CHECK ANSWERS. >, >, >,. p S - as S and p S as S -. Horizontal asmptote is = ; vertical asmptote is =.. e =, S S - e =, e - =, S S - e- =.6 Continuit The graphs of man functions encountered in this tet contain no holes, jumps, or reaks. For eample, if L = f t represents the length of a fish t ears after it is hatched, then the length of the fish changes graduall as t increases. Consequentl, the graph of L = f t contains no reaks (Figure.a). Some functions, however, do contain arupt changes in their values. Consider a parking meter that accepts onl quarters and each quarter us minutes of parking. Letting ct e the cost (in dollars) of parking for t minutes, the graph of c has reaks at integer multiples of minutes (Figure.). L. Length (in) L f (t) Cost (dollars)..7. c(t). t Time (r) (a) FIGURE. 6 t Time (min) () QUICK CHECK For what values of t in, 6 does the graph of = ct in Figure. have a discontinuit? Informall, we sa that a function f is continuous at a if the graph of f contains no holes or reaks at a (that is, if the graph near a can e drawn without lifting the pencil). If a function is not continuous at a, then a is a point of discontinuit. Continuit at a Point This informal description of continuit is sufficient for determining the continuit of simple functions, ut it is not precise enough to deal with more complicated functions such as h = sin if if =. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F.6 Continuit 7 It is difficult to determine whether the graph of h has a reak at ecause it oscillates rapidl as approaches (Figure.). We need a etter definition.. Is h continuous at?..... h() sin if if FIGURE. DEFINITION Continuit at a Point A function f is continuous at a if f = f a. If f is not continuous at a, then a is a point of discontinuit. Sa There is more to this definition than first appears. If f = f a, then f a and Sa f must oth eist, and the must e equal. The following checklist is helpful in Sa determining whether a function is continuous at a. f () Continuit Checklist In order for f to e continuous at a, the following three conditions must hold.. f a is defined a is in the domain of f.. Sa f eists.. Sa f = f a the value of f equals the it of f at a. 7 FIGURE. In Eample, the discontinuities at = and = are called removale discontinuities ecause the can e removed redefining the function at these points (in this case f = and f = ). The discontinuit at = is called a jump discontinuit. The discontinuit at = is called an infinite discontinuit. These terms are discussed in Eercises 9. If an item in the continuit checklist fails to hold, the function fails to e continuous at a. From this definition, we see that continuit has an important practical consequence: If f is continuous at a, then Sa f = f a, and direct sustitution ma e used to evaluate Sa f. EXAMPLE Points of discontinuit Use the graph of f in Figure. to identif values of on the interval, 7 at which f has a discontinuit. SOLUTION The function f has discontinuities at =,,, and ecause the graph contains holes or reaks at each of these locations. These claims are verified using the continuit checklist. f is not defined. f = and f =. Therefore, f and f eist ut are not equal. S S Copright Pearson Education, Inc.
or Use Onl in Pilot Program F 8 Chapter Limits f does not eist ecause the left-sided it f = differs from the S S - right-sided it f =. S + Neither S f nor f eists. Related Eercises 9 EXAMPLE Identifing discontinuities Determine whether the following functions are continuous at a. Justif each answer using the continuit checklist. a. f = + + ; a = -. g = + + ; a = - c. h = sin if ; a = if = SOLUTION a. The function f is not continuous at ecause f is undefined.. Because g is a rational function and the denominator is nonzero at, it follows Theorem. that S g = g = 7. Therefore, g is continuous at. c. B definition, h =. In Eercise of Section., we used the Squeeze Theorem to show that sin =. Therefore, h = h, which implies that h is S S continuous at. Related Eercises The following theorems make it easier to test various cominations of functions for continuit at a point. THEOREM.9 Continuit Rules If f and g are continuous at a, then the following functions are also continuous at a. Assume c is a constant and n 7 is an integer. a. f + g. f - g c. cf d. fg e. f >g, provided ga f. f n To prove the first result, note that if f and g are continuous at a, then Sa f = f a and Sa g = ga. From the it laws of Theorem., it follows that f + g = f a + ga. Sa Therefore, f + g is continuous at a. Similar arguments lead to the continuit of differences, products, quotients, and powers of continuous functions. The net theorem is a direct consequence of Theorem.9. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F.6 Continuit 9 f () 7 THEOREM. Polnomial and Rational Functions a. A polnomial function is continuous for all.. A rational function (a function of the form p, where p and q are polnomials) q is continuous for all for which q. EXAMPLE Appling the continuit theorems For what values of is the function f = - 7 + continuous? 6 SOLUTION a. Because f is rational, Theorem. implies it is continuous for all at which the denominator is nonzero. The denominator factors as - -, so it is zero at = and =. Therefore, f is continuous for all ecept = and = (Figure.). Related Eercises 6 FIGURE. Continuous everwhere ecept and The following theorem allows us to determine when a composition of two functions is continuous at a point. Its proof is informative and is outlined in Eercise. THEOREM. Continuit of Composite Functions at a Point If g is continuous at a and f is continuous at ga, then the composite function f g is continuous at a. QUICK CHECK Evaluate + 9 and + 9. S S How do these results illustrate that the order of a function evaluation and a it ma e switched for continuous functions? Theorem. is useful ecause it allows us to conclude that the composition of two continuous functions is continuous at a point. For eample, the composite function a - is continuous for all. The theorem also sas that under the stated conditions on f and g, the it of their composition is evaluated direct sustitution; that is, f g = f ga. Sa EXAMPLE Limit of a composition Evaluate a - + S 6 + +. SOLUTION The rational function - + 6 + is continuous for all ecause its + denominator is alwas positive (Theorem.). Therefore, a - + 6 + +, which is the composition of the continuous function f = and a continuous rational function, is continuous for all Theorem.. B direct sustitution, a - + S 6 + + = a - # + 6 + # + = =. Related Eercises 7 Closel related to Theorem. are two results dealing with its of composite functions; the are used frequentl in upcoming chapters. We present these two results one a more general version of the other in a single theorem. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits THEOREM. Limits of Composite Functions. If g is continuous at a and f is continuous at ga, then f g = f Sa g Sa.. If Sa g = L and f is continuous at L, then f g = f Sa g Sa. Proof: The first statement follows directl from Theorem., which states that f g = f ga. If g is continuous at a, then g = ga, and it follows that Sa Sa f g = f ga = f Sa Sa g. g Sa The proof of the second statement relies on the formal definition of a it, which is discussed in Section.7. Both statements of Theorem. justif interchanging the order of a it and a function evaluation. B the second statement, the inner function of the composition needn t e continuous at the point of interest, ut it must have a it at that point. EXAMPLE Limits of composite functions Evaluate the following its. a. S- - SOLUTION. S cos a - - a. We show later in this section that is continuous for Ú. The inner function of the composite function - is - and it is continuous and positive at -. B the first statement of Theorem., - = - = =. S- A S-. We show later in this section that cos is continuous at all points of its domain. The inner function of the composite function cos a - - is -, which is not - continuous at. However, e - S a - = - + S - Therefore, the second statement of Theorem., S cos a - - = cos a S = S + =. a - = cos -.6. - g Related Eercises Copright Pearson Education, Inc.
or Use Onl in Pilot Program F f () Continuous on [a, ) Continuit on an Interval.6 Continuit A function is continuous on an interval if it is continuous at ever point in that interval. Consider the functions f and g whose graphs are shown in Figure.6. Both these functions are continuous for all in a,, ut what aout the endpoints? To answer this question, we introduce the ideas of left-continuit and right-continuit. O a g() (a) Continuous on (a, ] DEFINITION Continuit at Endpoints A function f is continuous from the left (or left-continuous) at a if Sa - f = f a and f is continuous from the right (or right-continuous) at a if Sa + f = f a. Comining the definitions of left-continuous and right-continuous with the definition of continuit at a point, we define what it means for a function to e continuous on an interval. DEFINITION Continuit on an Interval O a FIGURE.6 () A function f is continuous on an interval I if it is continuous at all points of I. If I contains its endpoints, continuit on I means continuous from the right or left at the endpoints. To illustrate these definitions, consider again the functions in Figure.6. In Figure.6a, f is continuous from the right at a ecause f = f a; ut it is not continuous from Sa + the left at ecause f is not defined. Therefore, f is continuous on the interval a,. The ehavior of the function g in Figure.6 is the opposite: It is continuous from the left at, ut it is not continuous from the right at a. Therefore, g is continuous on a,. QUICK CHECK Modif the graphs of the functions f and g in Figure.6 to otain functions that are continuous on a,. f () Continuous on (, ) EXAMPLE 6 Intervals of continuit Determine the intervals of continuit for f = e + if + if 7. SOLUTION This piecewise function consists of two polnomials that descrie a paraola and a line (Figure.7). B Theorem., f is continuous for all. From its graph, it appears that f is left-continuous at. This oservation is verified noting that f = + =, S - S - Continuous on (, ] FIGURE.7 Left-continuous at which means that S - f = f. However, ecause f = + = f, S + S + we see that f is not right-continuous at. Therefore, we can also sa that f is continuous on -, and on,. Related Eercises Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits Functions Involving Roots Recall that Limit Law 7 of Theorem. states f n>m = Sa f Sa n>m, provided f Ú, for near a, if m is even and n>m is reduced. Therefore, if m is odd and f is continuous at a, then f n>m is continuous at a, ecause f n>m = Sa f Sa n>m = f a n>m. When m is even, the continuit of f n>m must e handled more carefull ecause this function is defined onl when f Ú. Eercise 9 of Section.7 estalishes an important fact: If f is continuous at a and f a 7, then f is positive for all values of in the domain sufficientl close to a. Comining this fact with Theorem. (the continuit of composite functions), it follows that f n>m is continuous at a provided f a 7. At points where f a =, the ehavior of f n>m varies. Often we find that f n>m is left- or right-continuous at that point, or it ma e continuous from oth sides. THEOREM. Continuit of Functions with Roots Assume that m and n are positive integers with no common factors. If m is an odd integer, then f n>m is continuous at all points at which f is continuous. If m is even, then f n>m is continuous at all points a at which f is continuous and f a 7. Continuous on [, ] Right-continuous at FIGURE.8 g() 9 Left-continuous at QUICK CHECK On what interval is f = > continuous? On what interval is f = > continuous? EXAMPLE 6 Continuit with roots For what values of are the following functions continuous? a. g = 9 -. f = - + > SOLUTION a. The graph of g is the upper half of the circle + = 9 (which can e verified solving + = 9 for ). From Figure.8, it appears that g is continuous on -,. To verif this fact, note that g involves an even root m =, n = in Theorem.). If - 6 6, then 9-7 and Theorem., g is continuous for all on -,. At the right endpoint, 9 - = = g Limit Law 7, which implies S - that g is left-continuous at. Similarl, g is right-continuous at - ecause 9 - = = g-. Therefore, g is continuous on -,. S - +. The polnomial - + is continuous for all Theorem.a. Because f involves an odd root (m =, n = in Theorem.), f is continuous for all. Related Eercises Continuit of Transcendental Functions The understanding of continuit that we have developed with algeraic functions ma now e applied to transcendental functions. Trigonometric Functions In Eample 8 of Section., we used the Squeeze Theorem to show that sin = and cos =. Because sin = and cos =, these S S Copright Pearson Education, Inc.
or Use Onl in Pilot Program F.6 Continuit its impl that sin and cos are continuous at. The graph of = sin (Figure.9) suggests that sin = sin a for an value of a, which means that sin is continuous Sa everwhere. The graph of = cos also indicates that cos is continuous for all. Eercise outlines a proof of these results. With these facts in hand, we appeal to Theorem.9e to discover that the remaining trigonometric functions are continuous on their domains. For eample, ecause sec = >cos, the secant function is continuous for all for which cos (for all ecept odd multiples of p>) (Figure.). Likewise, the tangent, cotangent, and cosecant functions are continuous at all points of their domains. sec sin sin a (a, sin a)... sin sin a w q q w FIGURE.9 a As a... FIGURE. sec is continuous at all points of its domain. f () Eponential Functions The continuit of eponential functions of the form f =, with 6 6 or 7, raises an important question. Consider the function f = (Figure.). Evaluating f is routine if is rational: = # # = 6; - = = 6 ; > = = 8; and -> =. (, ) Eponential functions are defined for all real numers and are continuous on (, ), as shown in Chapter 6 FIGURE. But what is meant when is an irrational numer, such as? In order for f = to e continuous for all real numers, it must also e defined when is an irrational numer. Providing a working definition for an epression such as requires mathematical results that don t appear until Chapter 6. Until then, we assume without proof that the domain of f = is the set of all real numers and that f is continuous at all points of its domain. Inverse Functions Suppose a function f is continuous and one-to-one on an interval I. Reflecting the graph of f through the line = generates the graph of f -. The reflection process introduces no discontinuities in the graph of f -, so it is plausile (and indeed, true) that f - is continuous on the interval corresponding to I. We state this fact without a formal proof. THEOREM. Continuit of Inverse Functions If a continuous function f has an inverse on an interval I, then its inverse f - is also continuous (on the interval consisting of the points f, where is in I). Because all the trigonometric functions are continuous on their domains, the are also continuous when their domains are restricted for the purpose of defining inverse functions. Therefore, Theorem., the inverse trigonometric functions are continuous at all points of their domains. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F Chapter Limits Logarithmic functions of the form f = log are continuous at all points of their domains for the same reason: The are inverses of eponential functions, which are one-to-one and continuous. Collecting all these facts together, we have the following theorem. THEOREM. Continuit of Transcendental Functions The following functions are continuous at all points of their domains. Trigonometric Inverse Trigonometric Eponential sin cos sin - cos - e tan cot tan - cot - Logarithmic sec csc sec - csc - log ln For each function listed in Theorem., we have Sa f = f a, provided a is in the domain of the function. This means that its involving these functions ma e evaluated direct sustitution at points in the domain. EXAMPLE 7 Limits involving transcendental functions Evaluate the following its after determining the continuit of the functions involved. a. S cos - cos - SOLUTION. S ln + tan - Limits like the one in Eample 7a are denoted / and are known as indeterminate forms, to e studied further in Section.7. QUICK CHECK Show that f = ln is right-continuous at =. a. Both cos - and cos - are continuous for all Theorems.9 and.. However, the ratio of these functions is continuous onl when cos -, which occurs when is not an integer multiple of p. Note that oth the numerator and denominator of cos - approach as S. To evaluate the it, we factor and simplif: cos - cos - S cos - = cos - cos + S cos - = S cos + (where cos - ma e canceled ecause it is nonzero as approaches ). The it on the right is now evaluated using direct sustitution: cos + = cos + =. S. B Theorem., ln is continuous on its domain,. However, ln 7 onl when 7, so Theorem. implies ln is continuous on,. At =, ln is right-continuous (Quick Check ). The domain of tan - is all real numers, and it is continuous on -,. Therefore, f = ln + tan - is continuous on,. Because the domain of f does not include points with 6, ln + tan - does not eist, which implies that ln + tan - S - S does not eist. Related Eercises 6 We close this section with an important theorem that has oth practical and theoretical uses. The Intermediate Value Theorem A common prolem in mathematics is finding solutions to equations of the form f = L. Before attempting to find values of satisfing this equation, it is worthwhile to determine whether a solution eists. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F.6 Continuit The eistence of solutions is often estalished using a result known as the Intermediate Value Theorem. Given a function f and a constant L, we assume L lies etween f a and f. The Intermediate Value Theorem sas that if f is continuous on a,, then the graph of f must cross the horizontal line = L at least once (Figure.). Although this theorem is easil illustrated, its proof goes eond the scope of this tet. Intermediate Value Theorem f () f () f (a) f () L L f (a) f () O a c O a c c c f () f is not continuous on [a, ]... f () FIGURE. In (a, ), there is at least one numer c such that f (c) L, where L is etween f (a) and f (). THEOREM.6 The Intermediate Value Theorem Suppose f is continuous on the interval a, and L is a numer strictl etween f a and f. Then there eists at least one numer c in a, satisfing f c = L. L f (a) O a FIGURE.... and there is no numer c in (a, ) such that f (c) L. QUICK CHECK 6 Does the equation f = + + = have a solution on the interval -,? Eplain. Amount of mone (dollars) after ears A(r) Interest rate that ields $ after ears..67. r Interest rate FIGURE. The importance of continuit in Theorem.6 is illustrated in Figure., where we see a function f that is not continuous on a,. For the value of L shown in the figure, there is no value of c in a, satisfing f c = L. The net eample illustrates a practical application of the Intermediate Value Theorem. EXAMPLE 8 Finding an interest rate Suppose ou invest $ in a special -ear savings account with a fied annual interest rate r, with monthl compounding. The amount of mone A in the account after ears (6 months) is Ar = a + r 6. Your goal is to have $ in the account after ears. a. Use the Intermediate Value Theorem to show there is a value of r in (,.8) that is, an interest rate etween % and 8% for which Ar =.. Use a graphing utilit to illustrate our eplanation in part (a), and then estimate the interest rate required to reach our goal. SOLUTION a. As a polnomial in r (of degree 6), Ar = a + r 6 is continuous for all r. Evaluating Ar at the endpoints of the interval,.8, we have A = and A.8 89.8. Therefore, A 6 6 A.8, and it follows, the Intermediate Value Theorem, that there is a value of r in,.8 for which Ar =.. The graphs of = Ar and the horizontal line = are shown in Figure.; it is evident that the intersect etween r = and r =.8. Solving Ar = algeraicall or using a root finder reveals that the curve and line intersect at r.67. Therefore, an interest rate of approimatel 6.7% is required for the investment to e worth $ after ears. Related Eercises 7 6 Copright Pearson Education, Inc.
or Use Onl in Pilot Program F 6 Chapter Limits SECTION.6 EXERCISES Review Questions. Which of the following functions are continuous for all values in their domain? Justif our answers. a. at = altitude of a skdiver t seconds after jumping from a plane. nt = numer of quarters needed to park in a metered parking space for t minutes c. Tt = temperature t minutes after midnight in Chicago on Januar d. pt = numer of points scored a asketall plaer after t minutes of a asketall game. Give the three conditions that must e satisfied a function to e continuous at a point.. What does it mean for a function to e continuous on an interval?. We informall descrie a function f to e continuous at a if its graph contains no holes or reaks at a. Eplain wh this is not an adequate definition of continuit.. Complete the following sentences. a. A function is continuous from the left at a if.. A function is continuous from the right at a if. 6. Descrie the points (if an) at which a rational function fails to e continuous. 7. What is the domain of f = e > and where is f continuous? 8. Eplain in words and pictures what the Intermediate Value Theorem sas. Basic Skills 9. Discontinuities from a graph Determine the points at which the following functions f have discontinuities. For each point, state the conditions in the continuit checklist that are violated. 9... f () f (). f () f (). Continuit at a point Determine whether the following functions are continuous at a. Use the continuit checklist to justif our answer.. f = + + ; a = +. f = + + ; a = - +. f = - ; a = 6. g = - ; a = - if 7. f = c - ; a = if = - + if 8. f = c - ; a = if = - 9. f = - 9 + ; a = + if -. f = c + ; a = - if = - 6. Continuit on intervals Use Theorem. to determine the intervals on which the following functions are continuous.. p = - +. g = - 6 + 7 + +. f = + 6 + 7-9. f = -. s = - + - 6. f t = t + t - 7. Limits of compositions Evaluate the following its and justif our answer. 7. 8-6 - 8. a S S - - 9. a + S +. a + S. Limits of composite functions Evaluate the following its and justif our answer. - - 8. SB -. ln a sin S.. tan t - ts t - a S 6 + - 8. Intervals of continuit Determine the intervals of continuit for the following functions.. The graph of Eercise 9 6. The graph of Eercise > Copright Pearson Education, Inc.
or Use Onl in Pilot Program F T 7. The graph of Eercise 8. The graph of Eercise 9. Intervals of continuit Let f = e + if Ú if 6. a. Use the continuit checklist to show that f is not continuous at.. Is f continuous from the left or right at? c. State the interval(s) of continuit.. Intervals of continuit Let f = e + + if if 7. a. Use the continuit checklist to show that f is not continuous at.. Is f continuous from the left or right at? c. State the interval(s) of continuit. 6. Functions with roots Determine the interval(s) on which the following functions are continuous. Be sure to consider right- and left-continuit at the endpoints.. f = - 6. g = -. f = - -. f t = t - >. f = - > 6. f z = z - > 7. Limits with roots Determine the following its and justif our answers. + 7. S A - 9. S + 7 8. - + - 9 S -. ts t + + t + 6. Continuit and its with transcendental functions Determine the interval(s) on which the following functions are continuous; then evaluate the given its.. f = csc ; f ; Sp>. f = e ; S f ; S + Sp - f f. f = + sin ; cos f ; f Sp> - Sp>. f = ln sin - ;. f = S - f e - e ; f ; f S - S + 6. f = e - e - ; f S 7. Intermediate Value Theorem and interest rates Suppose $ is invested in a savings account for ears ( months), with an annual interest rate of r, compounded monthl. The amount of mone in the account after ears is Ar = + r>. a. Use the Intermediate Value Theorem to show there is a value of r in (,.8) an interest rate etween % and 8% that allows ou to reach our savings goal of $7 in ears. T.6 Continuit 7. Use a graph to illustrate our eplanation in part (a); then approimate the interest rate required to reach our goal. 8. Intermediate Value Theorem and mortgage paments You are shopping for a $,, -ear (6-month) loan to u a house. The monthl pament is mr =,r> - + r> -6, where r is the annual interest rate. Suppose anks are currentl offering interest rates etween 6% and 8%. a. Use the Intermediate Value Theorem to show there is a value of r in (.6,.8) an interest rate etween 6% and 8% that allows ou to make monthl paments of $ per month.. Use a graph to illustrate our eplanation to part (a). Then determine the interest rate ou need for monthl paments of $. T 9 6. Appling the Intermediate Value Theorem a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval.. Use a graphing utilit to find all the solutions to the equation on the given interval. c. Illustrate our answers with an appropriate graph. 9. + - = ; -, 6. + + = ;, 6. - + = -; -, 6. - - + + = ;, 6. + e = ; -, 6. ln - = ;, e Further Eplorations 6. Eplain wh or wh not Determine whether the following statements are true and give an eplanation or countereample. a. If a function is left-continuous and right-continuous at a, then it is continuous at a.. If a function is continuous at a, then it is left-continuous and right-continuous at a. c. If a 6 and f a L f, then there is some value of c in a, for which f c = L. d. Suppose f is continuous on a,. Then there is a point c in a, such that f c = f a + f >. 66. Continuit of the asolute value function Prove that the asolute value function is continuous for all values of. (Hint: Using the definition of the asolute value function, compute and.) - + S S 67 7. Continuit of functions with asolute values Use the continuit of the asolute value function (Eercise 66) to determine the interval(s) on which the following functions are continuous. 67. f = + + - 8 68. g = ` - ` 69. h = ` - ` 7. h = + + + Copright Pearson Education, Inc.
or Use Onl in Pilot Program F 8 Chapter Limits 7 8. Miscellaneous its Evaluate the following its. cos + cos + 7. Sp cos + 7. Sp> sin - sin - cos - 7. S sin tan - 77. S 79. S - ln sin + 6 sin + 7. Sp> sin - + sin u - 7. us sin u - cos 76. S + sin cos t 78. ts e t 8. S + ln 8. Pitfalls using technolog The graph of the sawtooth function = - :;, where :; is the greatest integer function or floor function (Eercise 7, Section.), was otained using a graphing utilit (see figure). Identif an inaccuracies appearing in the graph and then plot an accurate graph hand. :;.. 8. Pitfalls using technolog Graph the function f = sin T using a graphing window of - p, p *,. a. Sketch a cop of the graph otained with our graphing device and descrie an inaccuracies appearing in the graph.. Sketch an accurate graph of the function. Is f continuous at? sin c. What is the value of S. 8. Sketching functions a. Sketch the graph of a function that is not continuous at, ut is defined at.. Sketch the graph of a function that is not continuous at, ut has a it at. 8. An unknown constant Determine the value of the constant a for which the function is continuous at -. 8. An unknown constant Let + + if - f = c + a if = - + if 6 g = a if = + if 7. a. Determine the value of a for which g is continuous from the left at.. Determine the value of a for which g is continuous from the right at. c. Is there a value of a for which g is continuous at? Eplain. T 86. Asmptotes of a function containing eponentials Let f = e + e e. Evaluate - e f, f, f, S - S + S- and f. Then give the horizontal and vertical asmptotes S of f. Plot f to verif our results. T 87. Asmptotes of a function containing eponentials Let f = e + e - e + e -. Evaluate f, f, and S S- f. Then give the horizontal and vertical asmptotes of f. S Plot f to verif our results. T 88 89. Appling the Intermediate Value Theorem Use the Intermediate Value Theorem to verif that the following equations have three solutions on the given interval. Use a graphing utilit to find the approimate roots. 88. + - + = ; -, 89. 7-87 + - = ;, Applications 9. Parking costs Determine the intervals of continuit for the parking cost function c introduced at the outset of this section (see figure). Consider t 6. Cost (dollars)...7.. c(t) 6 t Time (min) 9. Investment prolem Assume ou invest $ at the end of each ear for ears at an annual interest rate of r. The amount of mone in our account after ears is A = + r -. r Assume our goal is to have $ in our account after ears. a. Use the Intermediate Value Theorem to show that there is an interest rate r in the interval.,. etween % and % that allows ou to reach our financial goal.. Use a calculator to estimate the interest rate required to reach our financial goal. 9. Appling the Intermediate Value Theorem Suppose ou park our car at a trailhead in a national park and egin a -hr hike to a lake at 7 a.m. on a Frida morning. On Sunda morning, ou leave the lake at 7 a.m. and start the -hr hike ack to our car. Assume the lake is mi from our car. Let f t e our distance from the car t hours after 7 a.m. on Frida morning and let gt e our distance from the car t hours after 7 a.m. on Sunda morning. Copright Pearson Education, Inc.
or Use Onl in Pilot Program F.6 Continuit 9 a. Evaluate f, f, g, and g.. Let ht = f t - gt. Find h and h. c. Use the Intermediate Value Theorem to show that there is some point along the trail that ou will pass at eactl the same time of morning on oth das. 9. The monk and the mountain A monk set out from a monaster in the valle at dawn. He walked all da up a winding path, stopping for lunch and taking a nap along the wa. At dusk, he arrived at a temple on the mountaintop. The net da, the monk made the return walk to the valle, leaving the temple at dawn, walking the same path for the entire da, and arriving at the monaster in the evening. Must there e one point along the path that the monk occupied at the same time of da on oth the ascent and descent? (Hint: The question can e answered without the Intermediate Value Theorem.) (Source: Arthur Koestler, The Act of Creation.) Additional Eercises 9. Does continuit of f impl continuit of f? Let g = e if Ú - if 6. a. Write a formula for g.. Is g continuous at =? Eplain. c. Is g continuous at =? Eplain. d. For an function f, if f is continuous at a, does it necessaril follow that f is continuous at a? Eplain. 9 96. Classifing discontinuities The discontinuities in graphs (a) and () are removale discontinuities ecause the disappear if we define or redefine f at a so that f a = Sa f. The function in graph (c) has a jump discontinuit ecause left and right its eist at a ut are unequal. The discontinuit in graph (d) is an infinite discontinuit ecause the function has a vertical asmptote at a. O O Removale discontinuit a (a) Jump discontinuit a (c) f () f () O O Removale discontinuit a () a (d) f () f () 9. Is the discontinuit at a in graph (c) removale? Eplain. 96. Is the discontinuit at a in graph (d) removale? Eplain. Infinite discontinuit 97 98. Removale discontinuities Show that the following functions have a removale discontinuit at the given point. See Eercises 9 96. 97. f = - 7 + ; = - - if 98. g = c - if = ; = 99. Do removale discontinuities eist? Refer to Eercises 9 96. a. Does the function f = sin > have a removale discontinuit at =?. Does the function g = sin > have a removale discontinuit at =? T. Classifing discontinuities Classif the discontinuities in the following functions at the given points. See Eercises 9 96. -. f = - ; =. h = - + ; = and = -. Continuit of composite functions Prove Theorem.: If g is continuous at a and f is continuous at ga, then the composition f g is continuous at a. (Hint: Write the definition of continuit for f and g separatel; then comine them to form the definition of continuit for f g.). Continuit of compositions a. Find functions f and g such that each function is continuous at, ut f g is not continuous at.. Eplain wh eamples satisfing part (a) do not contradict Theorem... Violation of the Intermediate Value Theorem? Let f =. Then f - = - and f =. Therefore, f - 6 6 f, ut there is no value of c etween - and for which f c =. Does this fact violate the Intermediate Value Theorem? Eplain.. Continuit of sin and cos a. Use the identit sin a + h = sin a cos h + cos a sin h with the fact that sin = to prove that sin = sin a, S Sa there estalishing that sin is continuous for all. (Hint: Let h = - a so that = a + h and note that h S as S a.). Use the identit cos a + h = cos a cos h - sin a sin h with the fact that cos = to prove that cos = cos a. S Sa QUICK CHECK ANSWERS. t =,,. Both epressions have a value of, showing that f g = f g.. Fill Sa Sa in the endpoints.., ; -,. Note that S ln = ln = and f = ln =. + S + Because the it from the right and the value of the function at = are equal, the function is right-continuous at =. 6. The equation has a solution on the interval -, ecause f is continuous on -, and f - 6 6 f. Copright Pearson Education, Inc.