Chapter 5: Limits, Continuity, and Differentiability

Transcription

1 Chapter 5: Limits, Continuity, and Differentiability 63

2 Chapter 5 Overview: Limits, Continuity and Differentiability Derivatives and Integrals are the core practical aspects of Calculus. They were the first things investigated by Archimedes and developed by Liebnitz and Newton. The process involved eamining smaller and smaller pieces to get a sense of a progression toward a goal. This process was not formalized algebraically, though, at the time. The theoretical underpinnings of these operations were developed and formalized later by Bolzano, Weierstrauss and others. These core concepts in this area are Limits, Continuity and Differentiability. Derivatives and Integrals are defined in terms of limits. Continuity and Differentiability are important because almost every theorem in Calculus begins with the condition that the function is continuous and differentiable. The Limit of a function is the function value (y-value) epected by the trend (or sequence) of y-values yielded by a sequence of -values that approach the -value being investigated. In other words, the Limit is what the y-value should be for a given -value, even if the actual y-value does not eist. The limit was created/defined as an operation that would deal with y-values that were of an indeterminate form. Indeterminate Form of a Number--Defn: "A number for which further analysis is necessary to determine its value." Means: the number equals,,,,or other strange things. The formal definition is rather unwieldy and we will not deal with it in this course other that to show the Formal Definition and translate it: Formal Definition of a Limit Lim f () = L if and only if for every ε >, there eists δ > such that a if < a < δ, then f () L < ε. Formal Definition of a Limit means: When almost equals a, the limit almost equals y. 6

3 There are four kinds of Limits: Two-sided Limits (most often just referred to as Limits) One-sided Limits Infinite Limits Limits at Infinity Continuity basically means a function s graph has no breaks in it. The formal definition involved limits. Since all the families of functions investigated in PreCalculus are continuous in their domain, it is easier to look at when a curve is discontinuous rather than continuous. There are four kinds of discontinuity: Removable Discontinuity ( Lim a f () does eist) Essential Discontinuity ( Lim a f () does not eist) y y f(a) does not eist Point of Eclusion (POE) Vertical asymptotes y y f(a) eists Point of Displacement (POD) Jump Discontinuity 65

4 5.: Limits, L Hopital s Rule, and The Limit Definitions of a Derivative As mentioned in the intro to this chapter and last year, the limit was created/defined as an operation that would deal with y-values that were of an indeterminate form. lim f () is read "the limit, as approaches a, of f of." What the definition a means is, if is almost equal to a (the difference is smaller than some small numberδ ), f() is almost equal to L (the difference is smaller than some small numberε ). In fact, they are so close, we could round off and consider them equal. In practice, usually lim f() = f( a). In other words, the limit is the y for a given a = a--as long as y /. If y=/, we are allowed to factor and cancel the terms that gave the zeros. No matter how small the factors get, they cancel to as long as they do not quite equal /. E Find Lim ( + ) and 5 Lim ( + 3) Lim ( + ) = 5 + = 7 5 Lim ( + 3) = ( ) + 3 = 9 E Find Lim If = 5 here, 5 5 ( 5)( + 5), which is, would =. But with a Limit, is 5 only almost equal to 5, and, therefore, 5 =. So 5 Lim = ( + 5)( 5) Lim 5 5 Lim ( + 5) = 5 = = 66

5 As we can see from both the graph (in the given window) and the table, while no y-value eists for =5, the y-values of the points on either side of =5 show y should be. Basically, the limit allows us to factor and cancel before we substitute the number "a" for. OBJECTIVE Evaluate Limits algebraically. Evaluate Limits using L Hopital s Rule. Recognize and evaluate Limits which are derivatives. Use the nderiv function on the calculator to find numerical derivatives. E 3 Find lim Since if =, =, we must be able to factor and cancel this fraction. And, in fact, we know one of the factors in each must be otherwise the fraction would not yield zeros. So, lim + ( 6) lim + = + 3 ( )( + 5) + 6 = lim + 5 = 8 7, 67

6 E Find lim lim = 3 lim ( 5 + ) ( ) = lim = 35 5 = 7 E 5 Find lim Unlike the previous eamples, this fraction does not factor. Yet it must simplify, somehow, to eliminate the Indeterminate Number. If we multiply by conjugates to eliminate the radicals from the numerator: lim = lim = Lim = Lim ( ) + ( ) + + ( ) + = Lim + = + = or 68

7 One of the more powerful tools in Calculus for dealing with Indeterminate Forms and limits is called L Hopital s Rule. If f a g a = or L'Hopital's Rule, then Lim a = Lim a f g. f ' g' This rule allows us to evaluate limits of functions that do not factor, such as those that involve transcendental functions. E 5 (again) Find lim using L Hopital s Rule. Since, at =, lim = L Hopital s Rule is satisfied. lim = lim D D ( ) ( = ) lim = =, the condition for Obviously, this is a much faster process than the algebraic one. E 6 Demonstrate that lim sin =. Last year, we used this limit to prove the derivative of sin. but we never proved that the limit actually equals. This limit is relatively difficult to evaluate without L Hopital s Rule (and requires a complicated theorem called the Squeeze Theorem), but it is very easy with L H. 69

8 At =, sin =, L H applies. Lim sin = Lim cos = Obviously, this is not a proof, because of the circular reasoning. We used a derivative whose proof involved the limit that we were evaluating. But we are just interested in the process of applying L H here. E 7 Evaluate Lim + csc. This limit yields a different indeterminate form from before, namely,. Because the variable is in the base and the eponent, we might apply the logarithm rules before applying L Hopital s Rule. csc y = lim + ln y = lim csc ln + = lim L ' H = lim ln y = y = e The Limit Definitions of a Derivative ln + sin + cos As we recall from last year, the derivative was initially created as a function that would yield the slope of the tangent line. The slope formula for any line though two points is m = y y. When considering a tangent line, though, that equation has too many variables. We can simplify this some by realizing y = f. If we consider h to represent the horizontal distance between the points and realize that y = f, then the two points that form the secant line would be (, f ) and ( + h, f ( + h) ). 7

9 (+h, f(+h)) (, f()) f ( + h) f () f ( + h) f () Then the slope formula becomes m = or m =. + h h When h =, the points would merge and we would have the tangent line. h = gives m =, therefore, we can use the f( + h) f Lim and m = h tangent line lim h h will represent the slope of the tangent line. Lim h f ( + h) f () h is the derivative. Since we already know the derivative rules from the last chapter, we will not be using this very often. What we might see instead is the Numerical Derivative, which yields the slope of the tangent line at a specific point. There are two versions of the Numerical Derivative formula: The Numerical Derivative f '( a) = Lim a f () f (a) a or f '( a) = Lim h f (a + h) f (a) h Often, there are questions on the AP test that look like limit questions, but which are really questions about recognition of these formulas. 7

10 E 8 Evaluate Lim ( + h)3 () 3. h h We could FOIL the numerator out (using Pascal s Triangle for ease) and solve this limit algebraically: Lim ( + h)3 () 3 = Lim 8 +h + 3h + h 3 8 h h h h = Lim h + 3h + h 3 h h = Lim + h + h h = or we could apply L Hopital s Rule: Lim ( + h)3 () 3 3( + h) = Lim h h h = 3 = But the quickest way is to recognize that Lim ( + h)3 () 3 h h where f = 3. Of course, f ' = 3 and f ' = 3 =. is really f ', E 9 Evaluate ln( + ) h e h lim h Ln(e + h) Lim = d ( Ln ) = = h h d =e =e e 7

11 E Evaluate cos lim π π cos d lim = π d π = π ( π ) = sin = ( cos ) 73

12 5. Homework Set A Evaluate the following Limits.. Lim Lim cos Lim 3. Lim π cos cos 5cos 7 3. Lim tan. Lim ln. Lim 3 5. Lim e + 5. Lim ( cot ) 6. Lim csc ( + sec ) π 6. Lim e 7. Lim + e + 7. Lim Lim Lim ln 9. Lim + cot 9. Lim ln tan. Lim π sin 3sin sin. Lim Lim csc( 5)ln( ) 5. Lim π Lim + cos e 7

13 I. f 3 Lim f = 3 Evaluate the following Limits. II. does not eist. Graph of f III. Lim f 3. = Lim 3 + h h h. Lim sin π π 5. Lim e3+h e 3 h h 6. + sin Lim π + π 7. + Lim cos π + h h h 8. ln ln Lim 9. sin sin Lim 3 3. a Lim 3 3 a a 3. 6 a Lim 3 3 a a 3. Lim 3 + h h h Solve the following multiple choice problems 33. Given the graph of f below, tell which of the following are TRUE. y A. I only B. I and II C. II and III D. I, II, and III E. III only 75

14 I. f ( + h) Lim = h h 3. Given the graph of f f = II. ( Graph ) below, of f tell which of the following are TRUE. y III. Lim f = A. I only B. I and II C. II and III D. I, II, and III E. III only 35. If a, then Lim a a a is A.a B. 3a C. a 3 D. E. noneistent 36. If the average rate of change of a function f over the interval from = to = + h is given by e h sinh, then f ' = A. B. C. D. 3 E. 5. Homework Set B ln t. lim t tan t ln. lim e e tan 3. lim π + cos. lim π tan + cos 3. lim lim +ln e 76

15 ln 5. lim tan 5. lim lim t 3 e t 3 3 t 6. lim y π y + π cos y sin y 6. lim lim ln e e sin π e 7. lim sin sin 8. lim lim lim π cos π sin 9. lim e cos tan. lim 6 +. lim e ln e. lim t 3 t 3 9t t 5t + 6. lim csc ln + tan t. lim t ln e t 3. ln lim e e. lim sin sin 5. lim π tan π e 6. lim 5+h e 5 h h 6. lim e ( ln ) e 8. lim h + h 7( + h) + 6 ( 7 + 6) h h lim h h 77

16 3. ln lim e e h lim h h 78

17 5.: Continuity, One-sided Limits, and Infinite Limits In -, we said that Lim f () is basically what the y-value should be when = a, a even if a is not in the domain. For all of the families of functions that we studied last year, this was enough. But what if that y-value might be two different numbers? In a graph of a piece-wise defined function like this, It is not clear whether the y-value for = is or. As we can see from the table of values, -values less than have y-values that approach while -values greater than have y-values that approach : The algebraic ways to describe theses differences are one-sided limits. The symbols we use are: Lim a Lim a- f ()or Lim f () a + - f ()reads "the limit as approaches a from the left," while Lim f () reads a + "the limit as approaches a from the right." 79

18 In this eample, Lim - f () = and Lim f () =. The Lim f () does not eist, + because the one-sided limits do not equal each other. By the Lim f () does not eist, we mean there is not one REAL number that the limit equals. OBJECTIVE Evaluate one-sided limits graphically, numerically, and algebraically. Evaluate two-sided limits in terms of one-sided limits. Prove continuity or discontinuity of a given function. Interpret Vertical Asymptotes in terms of one-sided limits. E Does Lim - f () eist for f = +, if 3, if < For Lim f () to eist, Lim f ()must equal Lim f (). The domain states that any number less than - goes into 3 -. Therefore, Lim - f () = Lim 3 - = 3 = Similarly, numbers greater than - go into +, and Lim - f () = Lim ( +) + - = ( ) + = You can see that the two one-sided limits are not equal. Therefore, f () does not eist. Lim - We see in the graph that the two parts do not come together: 8

19 y There are two main topics where one-sided limits most often come into play: Continuity and Infinite Limits. A limit for a function only eists if the left and right limits are equal. Since this was not much of a concern for the functions in Precalculus, we pretty much ignored the fact, but in Calculus (both here and in college) teachers love to deal with this fact. lim a f eists if and only if lim f = lim f a a+ 8

20 y = g () E Evaluate each of the limits given the function pictured below. y a) b) c) lim g lim g + lim g d) g( ) e) lim g f) lim g + g) lim g h) g () i) lim g j) lim g + k) lim g l) g () a) e) i) - b) - f) j) - c) DNE g) k) - d) h) DNE l) - 8

21 CONTINUITY One of the main topics early in Calculus is CONTINUITY. It really is a simple concept, which, like the Limit, is made complicated by its mathematical definition. Let us take a look at the formal definition of continuous: Continuous--Defn: "A function f() is continuous at = a if and only if: i. f(a) eists*, ii. f () eists*, Lim a and iii. Lim a f () = f (a)." *By eists, we mean that it equals a real number. i) "f(a) eists" means a must be in the domain. ii) " Lim a f () eists" means Lim a- f () = Lim f (). a + iii) " Lim a f () = f (a)" should be self eplanatory. NB. All the families of functions which were eplored in PreCalculus are continuous in their domain. E 3 g = +, if >, if = 3, if < Is g() continuous at = -? To answer this question, we must check each part of the definition. i) Does g(-) eist? Yes, the middle line says that - is in the domain and it tells us that y = if = -. ii) iii) Does the Lim - Does Lim - g() eist? Yes, the two one-sided limits are equal. g() = g(-)? No. Lim - g()=, while g(-) = So g() is not continuous at = - because the limit does not equal the function. 83

22 E F = 5, if > +, if < Is F() continuous at =? Why not? F() is not continuous at = because is not in the domain. Notice neither inequality includes an equals sign. E 5 IfG = +, if > 3, if = 5, if <, is G() continuous at =? Why not? i) Does G() eist? Yes, the second line says that is in the domain (and it tells us that y = 3, if = ). ii) Does the Lim g() eist? No. The two-sided limit only eists if the - two one-sided limits are equal. But, Lim - G() = Lim - 5 = and Lim The two one-sided limits are not equal. Therefore, = 3 G() = Lim G() is not continuous at =, because Lim G()does not eist. INFINITE LIMITS The other situation where one-sided limits come into play is at vertical asymptotes. Here, the y-value goes to infinity (or negative infinity), which is why these limits are also called. E 6 Evaluate a) Lim and b) Lim - + In both these cases, we are considering a vertical asymptote. The limits, being y values, would be either positive or negative infinity, depending on if 8

23 the curve went up or down on that side of the asymptote. We can look at the limit algebraically (vs. graphically) though: a) Lim - = - = Note that the - is not from the left because the is from the left, but rather that - is negative for any values less than. b) Lim + = = + Note in this case that the numerator s sign effects the outcome (two negatives make a positive). Note: It is debatable whether the Infinite Limits eist or not. It depends on whether eist is defined as equal to a real number or not. Some books would say that Lim eists because there is one answer. It just happens to be a - transfinite number. Other books would say Lim because the answer is not a Real number. - There are certain Infinite Limits that we just need to know. does not eist (DNE) Lim + = Lim ( Ln ) = + Lim = Lim tan π + = Lim a = Lim tan π = The first two rules ( lim if a is negative. + a =, if a > and lim a =, if a > ) reverse signs All these rules are based on the asymptotic behavior of the functions. 85

24 y Here is the graph of y = ln. Here is the graph of y =. As we trace along the graph As we approaching trace along the from the graph right approaching side, the graph from heads the right down side, (to the ). graph heads up (to ). As we trace along the graph approaching from the left side, the graph heads down (to ). 3 3 y 3 86

25 6 y Here is the graph of y= tan on, π. [ ] As we trace along the graph approaching π from the right side, the graph heads down (to ). As we trace along the graph approaching π from π π π 3π π 5π the left side, the graph heads up (to ). 3π 6 87

26 I. f ' > Lim f I. II. Lim f = Lim f + f ' < Lim 5. Homework f II. f = III. Lim f III. Lim f Graph of f Graph of f. does The not graph eistof a function, f, is shown below. Which of the following statements are TRUE? y does not eist A. I only B. II only C. I, II, and III D. I and III E. II and III. The graph of a function, f, is shown below. Which of the following statements are TRUE? y A. I only B. II only C. I, II, and III D. I and III E. II and III 88

27 I. f is continuous at f = 3 f I. Lim f > Lim II. Lim f = Lim f II. Lim + f ( = Lim f ( 3 III. Lim f ( 3. ) does The not graph eist of a function, f, is shown below. Which of the following f f f f ( 3) III. Lim statements = Lim are TRUE? 3 3 Graph of f Graph of f y A. I only B. II only C. I, II, and III D. I and II E. II and III. The graph of a function, f, is shown below. Which of the following statements are TRUE? y 3 3 A. I only B. II only C. I, II, and III D. I and III E. II and III 89

28 I. I. Lim f f Lim = f = f Graph of f Graph of f II. II. Lim Lim f Lim f + f = Lim f + f III. Lim f f f f 5. fthe ( < ) Lim graph of a function, f, is shown below. Which of the following III. Lim does not eist statements are TRUE? y A. I only B. II only C. I, II, and III D. I and III E. II and III 6. The graph of a function, f, is shown below. Which of the following statements are TRUE? y A. I only B. I and III C. I, II, and III D. II only E. II and III Determine if each of the following functions is continuous at = a and use the definition to prove it. 9

29 7. f =, if >, if =, if < ; a = - 8. g = 5 ; a = - 9. h =, if ; a =, if =. k = +, if 3, if > ; a = 3. k = +, if 3, if > ; a = -. F = 3. H = +, if, if > +, if >, if =, if < ; a = - ; a = -. G = Cos, if π π, if > π ; a = π 5. f ( = tan 3) if 3 cos( 3) if > 3 ; a = 3 9

30 I. Lim f does not eist. Graph of f II. Lim f + III. Lim f = = 6. f = 3 if < ln + ; a = + 5 if > Evaluate the following Limits. ln 7. Lim + sin 8. Lim 7 9. Lim cot( 5) ln Solve the following multiple choice problems.. Given the graph of the function, f, below, which of the following statements are TRUE? y A. I only B. I and II C. II and III D. I, II, and III E. III only. Given the graph of the function, f, below, which of the following statements are TRUE? 9

31 I. Lim f does y = g not () eist. II. Lim f = Lim f + III. Lim f = 3 Graph of f y A. I only B. I and II C. II and III D. I, II, and III E. III only. Evaluate each of the limits given the function pictured below. y a) lim g b) lim g + c) lim g 93

32 d) g( ) e) lim g f) lim g + g) lim g h) g () i) lim g j) lim g + k) lim g l) g () 5. Homework Set B Evaluate the following Limits.. lim lim + ln tan π 3. lim ln ln( 5 t). lim t 5 t 3 ln( ) 7. lim + v 5. lim + v e v e 8. lim + +6 ln e 6. lim 3 tan + ( ) 9. lim tan ln t t. lim tan 5 5 ln 3. lim 3. lim ln 3. lim ln( ). lim + ( 5 ln ) 5. Evaluate each of the following for the graph of f (), shown below. y

33 a. lim f ( ) = b. lim + d. f ( ) = e. lim + f ( ) = c. lim f = f ( ) = f. lim f ( ) = g lim f = h. f = i. lim 3 + f ( ) = j. lim 3 m. lim + f ( ) = k. lim f 3 = l. f ( 3) = f ( ) = n. lim f ( ) = 6. Given the function f = a. lim f d. lim 6 b. lim f ( ) e. lim f find each of the following: c. lim f ( ) 6 + f ( ) 95

34 5.3: Differentiability and Smoothness The second (after Continuity) important underlying concept to Calculus is differentiability. Almost all theorems in Calculus begin with If a function is continuous and differentiable It is actually a very simple idea. Differentiability Defn: the derivative eists. F() is differentiable at a if and only if F (a) eists and is differentiable on [a, b] if and only F() is differentiable at every point in the interval. F() is differentiable at =a if and only if i. F() is continuous at =a, ii. Lim a- F'() and Lim F'() both eist, a + and iii. Lim a- F'() = Lim a + NB. All the families of functions studied in PreCaculus are Continuous and differentiable in their domains. OBJECTIVE Determine if a function is differentiable or not. Demonstrate understanding of the connections and differences between differentiability and continuity. There are two ways that a function would not be differentiable: I. The tangent line could be vertical, causing the slope to be infinite. E Is y = If y = 3 differentiable at =? 3, then dy d =. At =, dy 3 3 d = = dne. 96

35 y II. Just as the Limit does not eist if the two one-sided limits are not equal, a derivative would not eist if the two one-sided derivatives are not equal. That is, the slopes of the tangent lines to the left of a point are not equal to the tangent slopes to the right. A curve like this is called non-smooth. E Is the function represented by this curve differentiable at =? y We can see that the slope to the left of = is, while the slope to the right of = is. So this function is not differentiable at =. This is an eample of a curve that is not smooth. 97

36 E 3 Is G = Cos, if π π, if > π differentiable at = π? G' = Sin, if π π, if > π Lim π Lim π = = π π = π - G'() = Lim π Sin G'() = Lim π + π Clearly, these two one-sided derivatives are not equal. Therefore, G() is not differentiable at = π. y All three of these eamples show functions that are continuous, but not differentiable. Continuity does not insure differentiability. But the converse is true. If f() is differentiable, it MUST BE continuous and If f() is not continuous, it CANNOT BE differentiable 98

37 E Given that f = k +, if 3 + m, if > 3 is differentiable at =3, find m and k. f ' = If f() is differentiable at =3, then k k +, if 3 m, if > 3 3+ = m k = m. If f() is differentiable at =3, it is also continuous. Therefore, at =3, k + = + m k 3+ = + m3 k = 3m + Since both k = m and k = 3m + must be true, we can use linear combination or substitution to solve for k and m. k = 3 k + k = 3 k + 5 k = k = 8 5 m = 5 99

38 5.3 Homework. Determine if f = 3 is differentiable or not.. Determine if f = sin, if < ln, if is differentiable or not. 3. Determine if f = +, if ln, if > is differentiable or not.. Determine if f = + 5, if 3 +, if > is differentiable or not. 5. Given that f = m +, if k ln, if > is differentiable at =, find m and k. 6. Given that f = m 5, if k +, if > is differentiable at =, find m and k. 7. Given that f, if = ke 3 m, if > is differentiable at =, find m and k. 8. Given that f = m, if is differentiable at =, find m and k. k 3, if > 3

39 Graph of f Graph of f 9. At what values is f () not differentiable. y At what values is f () not differentiable. 3 y

40 Graph of f Graph of g. At what values is f () not differentiable. y At what value is function g not differentiable. y

41 Graph of f 3. Use the graph of f below to select the correct answer from the choices below. y = f ( 3) A. lim 3 f B. f is not continuous at = 3 C. f is differentiable at = 3 D. f ' E. lim 3 f < f ' does not eist. Use the graph of f below to select the correct answer from the choices below. y A. f has no etremes B. f is continuous at = C. f is differentiable for, E. f is concave up for (,) D. f has a relative maimum at 33

42 5.: Limits at Infinity and End Behavior Vocabulary Limit at Infinity Defn: the y-value when approaches infinity or negative infinity Infinite Limit Defn: a limit where y approaches infinity End Behavior Defn: The graphical interpretation of a limit at infinity OBJECTIVE Evaluate Limits at infinity. Interpret Limits at infinity in terms of end behavior of the graph. Evaluating limits at infinity for algebraic functions relies on one fact: Lim ± = This fact is what led to our general rule about horizontal asymptotes. E Evaluate Lim We already know, from last year, that this function, y = horizontal asymptote at y = 3. Why is this true? Because 3 + 3, has a 3

43 Lim = Lim = + 3 Lim = Lim 3 + Lim 3 = 3 + = E Find lim e and lim e and interpret these limits in terms of the end behavior of y = e. lim e = e =. This means that the right end (+ ) goes up. lim e = e = e = =. This means that, on the left (- ), y = e has a horizontal asymptote at y =. **NB. The synta we used here treating as if it were a real number and plugging it in is incorrect. Though it works as a practical approach, writing this on the AP test will lose you points for process. In particular, this will come up with Improper Integrals. We will discuss it again in Chapter 7. There are certain Limits at Infinity (which come from our study of end behavior) that we just need to know. Lim Tan = π or Lim Tan = π Lim e = and For y ' = ky( A y) and its solution y = Lim e = A + Be, kt Lim f = A These last two equations are the Logistic Growth equations. We will discuss them further in a later chapter. 35

44 E 3 Evaluate lim Tan ( 3 ) lim Tan 3 = Tan lim 3 = Tan = π Limits at Infinity often involve ratios and L Hospital s Rule can be applied. But some math teachers consider this to be using a cannon to kill a fly. It is easier to just know the relative growth rates of functions i.e., which families of functions grow faster than which. The End Behavior studied in PreCalculus gives a good sense of order. The Hierarchy of Functions. Logs grow the slowest.. Polynomials, Rationals and Radicals grow faster than logs and the degree of the End Behavior Model (EBM) determines which algebraic function grows fastest. For eample, y = grows more slowly than y =. 3. The trig inverses fall in between the algebraic functions at the value of their respective horizontal asymptotes.. Eponential functions grow faster than the others. (In BC Calculus, we will see the factorial function, y = n!, grows the fastest.) 5. The fastest growing function in the combination determines the end behavior, just as the highest degree term did among the algebraics. 36

45 E Evaluate (a) lim 3 e 95 and (b) lim 3 ln (a) Repeated iterations of L Hospital s Rule will give the same result, but, because eponential grow faster than polynomials, lim 3 e 95 = (b) Since 3 essentially has a degree of 3, the denominator has a higher degree and lim 3 ln =. EX 6 Use the concept of the Hierarchy of Functions to find the end behavior of: (a) y = e and (b) y = ln. (a) For y = e, the eponential y = e determines the end behavior. So the right end goes up and the left end has a horizontal asymptote at y =. (b) We still must consider the domain, first. There is no end behavior on left, because the domain is y (, ). For y = ln, y = dominates, so, on the right, the curve goes up. 37

46 5. Homework Set A Evaluate the Limit.. lim lim e 3. lim ( + e )Tan ( +). lim cos 5. lim lim ln Tan e + 7, lim sin ln 8. lim Tan lim Tan Tan lim e +. lim Cot e + Use limits to determine the end behavior of the following functions.. f = ( e +) ( Tan 3 ) 3. f = cos. f = ln 5. f = ln 38

47 5. Homework Set B. lim e. lim lim e 5. lim e lim tan + e 6. lim tan 7. lim tan 8. lim sin tan 5 + ( ) 9. lim e t t t +. lim y y + 5 y 3 y. lim lim 5+ e lim e 6. lim tan 5. lim lim y 6 5y y 8y + e 5 8. lim 5 +. limcos tan e t t t + e 7. limsin tan 5 t 5 + ( ) 9. lim e tan + + t ( 6 5y )e 3. lim y y 8y +. lim tan 3 3. lim e tan + + t 5+ 39

48 5.5 Type I Improper Integrals We have only looked into definite integrals over finite regions or integrals on bounded functions. Integrals over an infinite region are called Improper Integrals. Since an integral is a sum, it should make sense that unbounded integrals (integrals over an infinite area) might still be finite. That is, Improper Integrals can be Convergent or Divergent. There are two types of improper integrals: Type I: Definite integrals over an infinite interval (there s an infinity in one of the limits of integration) -- d Type II: Integrals over an unbounded function (i.e., integrating over a vertical asymptote) -- d In this section, we will only look at Type I Improper Integrals. Though the region is infinite in the -direction, the area might still be finite. Vocabulary Convergent Integral--Defn: an improper integral that has a total. Divergent Integral --Defn: an improper integral that does not have a total. Objectives Evaluate Type I Improper Integrals. Type I Improper Integral of come in three forms, based on the boundaries. Either the upper boundary, the lower boundary, or both boundaries are infinite. b b (a) If f d eists for every number b a, then f d a = lim f a b d, a provided this limit eists as a finite number. 3

49 b (b) If f d eists for every number a b, then a b b f d = lim f a d, provided this limit eists as a finite number. a c (c) If both f d and f d c are convergent, then c f d = f d + f d, where c is in ( a,b). c You can choose any c you want so choose an easy c, like or. If either of the two integrals on the right diverge, then the entire integral will diverge. The improper integrals are said to be convergent if these limits eist as a finite number and divergent if ANY of the limits do not eist. Steps to Solving an Improper Integral:. Recognize the integral as improper.. Rewrite the improper integral using limits. 3. Evaluate the integral, saving the limit until the very end.. When you are finished with the integral, you will have an epression containing the variable you are taking the limit of evaluate this limit. 5. If your limit gives you a number, the integral converges to that number. If your limit is undefined or infinite, the integral diverges. E Determine whether d is convergent or divergent. d = lim d Write the improper integral using limits b b b = lim d Convert to a power so we can integrate b b = lim b = lim b b ( ) Use the FTC Keep using the FTC = lim b b + Even more of the FTC = Because we got a number we would say our integral converges. 3

50 d converges to. E Determine whether d is convergent or divergent. b d = lim b d Write the improper integral using limits b = lim d b Evaluate the integral, leaving the limit alone b = lim ln b Why don t we need the absolute values? = lim lnb b ( ln ) Evaluate the limit = So, Note: Tuck this fact away d d diverges but is divergent. d converges. E 3 Determine whether + d is convergent or divergent. + d = + d + + d b = lim a + d + lim b + d a Let u = + du = d = lim a = =a du u + lim b a + = lim ln + a =b = lim b du u 3 b ln( + )

51 = lim a = + ln ln + a + lim b ln ln + b + is another one of those indeterminate forms in Calculus. And in terms of improper integrals, it means our integral is divergent because one of the limits is divergent. It does not matter that both are divergent. It only takes one. Really, you only would need to evaluate one integral at a time. If the first integral divergent, then the entire integral will diverge. If the first integral is convergent, evaluate the second one. The second integral also must converge for the entire integral to do converge. 33

52 5.5 Homework. ( 3 +) d. 3 w dw 3. e - d. y 3 dy 5. e - d 6. cosθ dθ 7. Ln d 8. t + dt 9. 5 d. + 9 d 3. e ln d. Which of the following improper integrals converge? I. d II. + d III. ln d (a) I only (b) II only (c) I and II only (d) II and III only (e) I and III only 3. u 3 du (a) (b) 5 (c) (d) - (e) DNE 3

53 5.6 Type II Improper Integrals Like Type I, Type II Improper Integrals occur in three situations, depending on the vertical asymptote. Either the Asymptote is at the beginning, the end or in the middle of the interval defined by the boundaries. If f is a continuous function on a,b) and is discontinuous at b, then b t f d = lim f a t b d, if this limit eists as a finite number. a If f is a continuous function on ( a,b and is discontinuous at a, then b b f d = lim f a t a + d, if this limit eists as a finite number. t c b If f has a discontinuity at c, where a < c < b, and both f d and f d a are c b c b convergent, then f d = f d + f d. a a c Objectives Determine convergence or divergence of a Type II Improper Integral. E Determine whether d is convergent or divergent. d b = lim Write the improper integral using limits b = lim b d b d Evaluate the integral, leaving the limit alone b = lim b ( b ) = lim b = lim b b Evaluate the limit = dne This integral diverges 35

54 E Determine whether d is convergent or divergent. d = lim a + d Write the improper integral using limits = lim a + a d Evaluate the integral, leaving the limit alone a = lim a + a = lim a + a Evaluate the limit = This integral converges to E 3 Determine whether 5 d is convergent or divergent. One might think this is an easy problem: 5 d = 5 d = Ln 5 = Ln 3 Ln 5 = Ln 3 5 This is a classic eample of what goes wrong if the integral is not recognized as improper. The answer Ln 3 5 is wrong!!! We need to split this into two integrals split the integral where the function has a discontinuity, in this case at = 5. 36

55 5 d 5 = 5 d + = lim b 5 = lim b 5 b 5 5 d 5 d + lim a 5 ln 5 b + lim + a 5 + a 5 d ln 5 a = lim ln b 5 ln5 b 5 + lim ln3 ln a 5 a 5 + lim b 5 ln b 5 = dne, therefore, 5 d is divergent. 37

56 5.6 Homework Determine whether each integral is convergent or divergent. Evaluate those that are convergent d. 3 + d π 3. csc t dt. 3 d π 5. sec d d 7. Which of the following improper integrals converge? I. d II. + d III. ln e d (a) I only (b) II only (c) I and II only (d) I and III only (e) I, II, and III only 8. u du (a) (b) (c) (d) (e) DNE 38

57 9. Which of the following improper integrals diverge? I. e d II. d III. d (a) I only (b) II only (c) III only (d) II and III only (e) I, II, and III 39

58 Limit and Continuity Test. The function f is differentiable at = b. Which of the following statements could be false? (a) lim b f eists (b) lim f b = f ( b) (c) lim b (d) lim b f ' ( ) = lim b + f ( ) = lim b + f ' ( ) (e) None of these f ( ) + k for < 5. The function f is defined for all Reals such that f = 5sin π for 5. For which value of k will the function be continuous throughout its domain? (a) (b) (c) 3 (d) (e) None of these 3. Which of the following improper integrals converge? II. e d II. d III. d (a) I only (b) III only (c) I and II only (d) II and III only (e) I and III only. If f I. lim 3 f = + for 3, which of the following statements are true? 7 for > 3 eists II. f is continuous at = 3 III. f is differentiable at = 3 (a) None (b) I only (c) II only (d) I and II only (e) I, II, and III 3

59 5. Which of the following functions is differentiable at =? (a) f = + (b) f = (c) f = sin for for = (d) f = for for = (e) f = cos for < sin for 6. lim h sin π + h = h (a) π (b) π (c) (d) π (e) DNE 7. lim sint dt 3 (a) (b) (c) 3 (d) 3 (e) DNE 8. lim = (a) (b) 3 (c) 3 (d) 3 (e) DNE 9. u 5 du (a) (b) 5 (c) (d) - (e) DNE 3

60 y f f I. lim dne f II. lim f = () III. lim. The graph of a function f is given below. Which of the following f dne statements are true? (A) I only (B) II only (C) I and II (D) I, II, and III (E) II and III. The graph of a function is shown below. 6 5 Which of the following statement(s) is (are) true? - I. lim II. lim + III. f lim eists eists eists 3

61 (a) I only (b) II only (c) I and II only (d) I and III only (e) I, II, and I y At which -value is f (graphed above) differentiable but not continuous? (a) (b) (c) (d) (e) nowhere 3. f = sin π, if <, if = Ln, if > a) Is f continuous? Why/Why not? b) Is f differentiable? Why/Why not? 33

62 . 3 ( 3 + ) 3 d 5. d 5. Given the graph of f () below, find the values of the following: a. Lim f () b. Lim f () c. Lim f () d. f ( 3 ) e. Lim f () f. Lim f () g. Lim + f () h. f ( ) i. Lim f () j. Lim f () k. Lim + m. Lim f () n. Lim f () o. Lim f () l. f f () p. f ( 3) 3

63 5. Homework e e a 3. 3 a C 3. D 35. C 36. E 5. Homework. E. B 3. D. D 5. C 6. B 7. Lim DNE 8. G(-) DNE 35

64 9. Continuous. Continuous. Lim DNE. Continuous 3. H(-) not equal to the LIM. Continuous 5. Continuous 6. dne 7. dne A. C a) lim g = ; b) d) g( ) = e) h) g () = D.N.E. i) l) g () = lim g = c) + lim g = f) lim g = j) + lim g = D.N.E. lim g = g) lim g = lim g = k) + lim g = 5.3 Homework. Not differentiable, dy d dne.. Not continuous. 3. Not differentiable, f '( ) f '( + ).. Differentiable 5. m =, k = 6. m = 6, k = 3 7. m = 6, k = 3 8. m = 3, k = 3 9. Not differentiable. Not differentiable. Not differentiable. Differentiable 3. B. B 36

65 5. Homework π. ln π π 9. π. π.. HA y = π on Left, Up on the left 3. HA at y =. HA at y = on Right 5. HA at y = on Left 5.5 Homework.. Diverges 3.. Diverges Diverges 7. Diverges 8. π 9. ln5. π 3.. A 3. E 5.6 Homework. 3. Diverges 3. Diverges. Diverges 5. Diverges 6. Diverges 7. D 8. E 9. E 37