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Unformatted text preview: S E C T I O N 2 . 3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S  99 37. (a) Evaluate the function f x x2 2 x 1000 for x 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of lim x 2 2x 1000 the origin several times. Comment on the behavior of this function.
40. In the theory of relativity, the mass of a particle with velocity v is xl0 (b) Evaluate f x for x 0.001. Guess again.
38. (a) Evaluate h x 0.04, 0.02, 0.01, 0.005, 0.003, and x x 3 for x 1, 0.5, 0.1, 0.05, m m0 s1
v2 c2 tan x ; tan x x . x3 (c) Evaluate h x for successively smaller values of x until you ﬁnally reach a value of 0 for h x . Are you still conﬁdent that your guess in part (b) is correct? Explain why you eventually obtained a value of 0. (In Section 4.4 a method for evaluating the limit will be explained.) (d) Graph the function h in the viewing rectangle 1, 1 by 0, 1 . Then zoom in toward the point where the graph crosses the yaxis to estimate the limit of h x as x approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c). (b) Guess the value of lim
xl0 0.01, and 0.005. where m 0 is the mass of the particle at rest and c is the speed of light. What happens as v l c ? ; 41. Use a graph to estimate the equations of all the vertical
asymptotes of the curve y tan 2 sin x x Then ﬁnd the exact equations of these asymptotes. ; 42. (a) Use numerical and graphical evidence to guess the value
of the limit lim x3 sx 1 1 xl1 ; 39. Graph the function f x
viewing rectangle sin x of Example 4 in the 1, 1 by 1, 1 . Then zoom in toward (b) How close to 1 does x have to be to ensure that the function in part (a) is within a distance 0.5 of its limit? 2.3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S In Section 2.2 we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits.
LIMIT LAWS Suppose that c is a constant and the limits
xla lim f x and xla lim t x exist. Then
1. lim f x
xla tx tx
xla xla lim f x xla lim t x 2. lim f x
xla xla lim f x xla lim t x 3. lim c f x
xla c lim f x
xla 4. lim f x t x
xla lim f x xla lim t x 5. lim xla fx tx xla lim f x
xla lim t x if lim t x
xla 0 100  C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S These ﬁve laws can be stated verbally as follows:
S U M L AW D I F F E R E N C E L AW CONSTANT MULTIPLE LAW P R O D U C T L AW QUOTIENT LAW 1. The limit of a sum is the sum of the limits. 2. The limit of a difference is the difference of the limits. 3. The limit of a constant times a function is the constant times the limit of the function. 4. The limit of a product is the product of the limits. 5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). It is easy to believe that these properties are true. For instance, if f x is close to L and t x is close to M, it is reasonable to conclude that f x t x is close to L M. This gives us an intuitive basis for believing that Law 1 is true. In Section 2.4 we give a precise definition of a limit and use it to prove this law. The proofs of the remaining laws are given in Appendix F. y 1 0 1 f x EXAMPLE 1 Use the Limit Laws and the graphs of f and t in Figure 1 to evaluate the following limits, if they exist. fx (a) lim f x (b) lim f x t x (c) lim 5t x xl 2 xl1 xl2 t x g SOLUTION (a) From the graphs of f and t we see that
xl 2 FIGURE 1 lim f x 1 and xl 2 lim t x 1 Therefore, we have
xl 2 lim fx 5t x xl 2 lim f x lim f x 5 1 xl 2 lim 5t x
xl 2 (by Law 1) (by Law 3) xl 2 5 lim t x 4 1 (b) We see that lim x l 1 f x right limits are different:
xl1 2. But lim x l 1 t x does not exist because the left and 2 lim t x 1 lim t x xl1 So we can’t use Law 4 for the desired limit. But we can use Law 4 for the onesided limits:
xl1 lim f xtx 2 2 4 xl1 lim f xtx 2 1 2 The left and right limits aren’t equal, so lim x l 1 f x t x does not exist. (c) The graphs show that
xl2 lim f x 1.4 and xl2 lim t x 0 Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. M S E C T I O N 2 . 3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S  101 If we use the Product Law repeatedly with t x
POWER LAW 6. lim f x
x la f x , we obtain the following law. n [ lim f x ]
x la n where n is a positive integer In applying these six limit laws, we need to use two special limits:
7. lim c
xla c 8. lim x
xla a These limits are obvious from an intuitive point of view (state them in words or draw graphs of y c and y x), but proofs based on the precise deﬁnition are requested in the exercises for Section 2.4. If we now put f x x in Law 6 and use Law 8, we get another useful special limit.
9. lim x n
xla an where n is a positive integer A similar limit holds for roots as follows. (For square roots the proof is outlined in Exercise 37 in Section 2.4.)
n 10. lim sx n sa xla where n is a positive integer 0.) (If n is even, we assume that a More generally, we have the following law, which is proved as a consequence of Law 10 in Section 2.5.
n 11. lim sf x) ROOT LAW x la s lim f x la
n x) where n is a positive integer
x la [If n is even, we assume that lim f x 0. EXAMPLE 2 Evaluate the following limits and justify each step. (a) lim 2 x 2
x l5 3x 4 (b) lim x3 xl 2 2x 2 1 5 3x lim 4
x l5 SOLUTION (a) lim 2 x 2
x l5 3x 4 lim 2 x 2
x l5 lim 3x
x l5 (by Laws 2 and 1) (by 3) (by 9, 8, and 7) 2 lim x 2
x l5 3 lim x
x l5 lim 4
x l5 2 52 39 35 4 102  C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S N E W TO N A N D L I M I T S Isaac Newton was born on Christmas Day in 1642, the year of Galileo’s death. When he entered Cambridge University in 1661 Newton didn’t know much mathematics, but he learned quickly by reading Euclid and Descartes and by attending the lectures of Isaac Barrow. Cambridge was closed because of the plague in 1665 and 1666, and Newton returned home to reﬂect on what he had learned. Those two years were amazingly productive for at that time he made four of his major discoveries: (1) his representation of functions as sums of inﬁnite series, including the binomial theorem; (2) his work on differential and integral calculus; (3) his laws of motion and law of universal gravitation; and (4) his prism experiments on the nature of light and color. Because of a fear of controversy and criticism, he was reluctant to publish his discoveries and it wasn’t until 1687, at the urging of the astronomer Halley, that Newton published Principia Mathematica. In this work, the greatest scientiﬁc treatise ever written, Newton set forth his version of calculus and used it to investigate mechanics, ﬂuid dynamics, and wave motion, and to explain the motion of planets and comets. The beginnings of calculus are found in the calculations of areas and volumes by ancient Greek scholars such as Eudoxus and Archimedes. Although aspects of the idea of a limit are implicit in their “method of exhaustion,” Eudoxus and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians such as Cavalieri, Fermat, and Barrow, the immediate precursors of Newton in the development of calculus, did not actually use limits. It was Isaac Newton who was the ﬁrst to talk explicitly about limits. He explained that the main idea behind limits is that quantities “approach nearer than by any given difference.” Newton stated that the limit was the basic concept in calculus, but it was left to later mathematicians like Cauchy to clarify his ideas about limits. (b) We start by using Law 5, but its use is fully justiﬁed only at the ﬁnal stage when we see that the limits of the numerator and denominator exist and the limit of the denominator is not 0. lim x 3 2 x 2 1 x 3 2x 2 1 xl 2 (by Law 5) lim xl 2 5 3x lim 5 3x
xl 2 xl 2 lim x 3
xl 2 2 lim x 2
xl 2 xl 2 xl 2 lim 1
(by 1, 2, and 3) lim 5 3 lim x 1 2 1 11
N OT E 3 5 2 22 32 (by 9, 8, and 7)
M If we let f x 2 x 2 3x 4, then f 5 39. In other words, we would have gotten the correct answer in Example 2(a) by substituting 5 for x. Similarly, direct substitution provides the correct answer in part (b). The functions in Example 2 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 53 and 54). We state this fact as follows.
DIRECT SUBSTITUTION PROPERTY If f is a polynomial or a rational function and a is in the domain of f , then
xla lim f x fa Functions with the Direct Substitution Property are called continuous at a and will be studied in Section 2.5. However, not all limits can be evaluated by direct substitution, as the following examples show.
EXAMPLE 3 Find lim xl1 x2 x 1 . 1 S O L U T I O N Let f x x 2 1 x 1 . We can’t ﬁnd the limit by substituting x 1 because f 1 isn’t deﬁned. Nor can we apply the Quotient Law, because the limit of the denominator is 0. Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares: x2 x 1 1 x 1x x1 1 The numerator and denominator have a common factor of x 1. When we take the limit as x approaches 1, we have x 1 and so x 1 0. Therefore we can cancel the common factor and compute the limit as follows: lim x2 x 1 1 lim x 1x x1 1 lim x 1 1 1 2 xl1 xl1 xl1 The limit in this example arose in Section 2.1 when we were trying to ﬁnd the tangent to the parabola y x 2 at the point 1, 1 . M
N OT E fx In Example 3 we were able to compute the limit by replacing the given function x 2 1 x 1 by a simpler function, t x x 1, with the same limit. This is S E C T I O N 2 . 3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S  103 y 3 2 1 0 y 3 2 1 0 1 2 3 x 1 2 3 x y=ƒ valid because f x t x except when x 1, and in computing a limit as x approaches 1 we don’t consider what happens when x is actually equal to 1. In general, we have the following useful fact. If f x t x when x a, then lim f x
xla xla lim t x , provided the limits exist. EXAMPLE 4 Find lim t x where
x l1 y=© tx
S O L U T I O N Here t is deﬁned at x x 1 if x if x 1 1 , but the value of a limit as x approaches 1 and t 1 1 does not depend on the value of the function at 1. Since t x x 1 for x 1, we have M lim t x lim x 1 2
xl1 xl1 The graphs of the functions f (from Example 3) and g (from Example 4) FIGURE 2 x Note that the values of the functions in Examples 3 and 4 are identical except when 1 (see Figure 2) and so they have the same limit as x approaches 1. Evaluate lim 3 h2 h 9 . 3 h2 h 9 0 since F 0 is V EXAMPLE 5 hl0 S O L U T I O N If we deﬁne Fh then, as in Example 3, we can’t compute lim h l 0 F h by letting h undeﬁned. But if we simplify F h algebraically, we ﬁnd that Fh 9 6h h 3 h2 h 3 . 9 h2 9 6h h h2 6 h (Recall that we consider only h
hl0 0 when letting h approach 0.) Thus
hl0 lim lim 6 h 6 M EXAMPLE 6 Find lim
tl0 st 2 9 t2 S O L U T I O N We can’t apply the Quotient Law immediately, since the limit of the denomi nator is 0. Here the preliminary algebra consists of rationalizing the numerator: lim
tl0 st 2 9 t
2 3 lim
tl0 st 2 9 t
2 3 st 2 st 2 9 3) 9 9 lim
tl0 2 3 3 t (st 1 9 t2
2 lim
tl0 t2 9 t (st 2 9
2 9 3 3) 1 3 3 1 6
M lim
tl0 st 2 1 9 3 s lim t 2
tl0 This calculation conﬁrms the guess that we made in Example 2 in Section 2.2. 104  C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S Some limits are best calculated by ﬁrst ﬁnding the left and righthand limits. The following theorem is a reminder of what we discovered in Section 2.2. It says that a twosided limit exists if and only if both of the onesided limits exist and are equal.
1 THEOREM xla lim f x L if and only if x la lim f x L x la lim f x When computing onesided limits, we use the fact that the Limit Laws also hold for onesided limits.
N The result of Example 7 looks plausible from Figure 3. y EXAMPLE 7 Show that lim x
xl0 0. S O L U T I O N Recall that
y= x  x Since x x for x 0, we have
x l0 x if x x if x 0 0 0 x lim x x l0 lim x 0 FIGURE 3 For x 0 we have x x and so
xl0 lim x xl0 lim x 0 Therefore, by Theorem 1,
xl0 lim x 0 M V EXAMPLE 8 y 1 0 x Prove that lim lim lim xl0 x does not exist. x x x x x lim lim x x x x lim 1 lim 1 1 1 x y= x SOLUTION x l0 x l0 x l0 _1 x l0 x l0 x l0 FIGURE 4 Since the right and lefthand limits are different, it follows from Theorem 1 that lim x l 0 x x does not exist. The graph of the function f x x x is shown in Figure 4 and supports the onesided limits that we found.
EXAMPLE 9 If M fx determine whether lim x l 4 f x exists.
S O L U T I O N Since f x
N sx 4 8 2x if x if x 4 4 It is shown in Example 3 in Section 2.4 that lim x l 0 sx sx
x l4 4 for x
x l4 4, we have lim s x 4 s4 4 0 0. lim f x S E C T I O N 2 . 3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S  105 y Since f x 8 2 x for x
x l4 4, we have
x l4 lim f x lim 8 2x 8 24 0 0 4 x The right and lefthand limits are equal. Thus the limit exists and
xl4 FIGURE 5 lim f x 0
M The graph of f is shown in Figure 5.
Other notations for x are x and x . The greatest integer function is sometimes called the ﬂoor function.
N EXAMPLE 10 The greatest integer function is deﬁned by x y 4 3 2 1 0 1 2 3 4 5 x that is less than or equal to x. (For instance, 4 4, 4.8 1 1.) Show that lim x l 3 x does not exist. 2 for 3 x 4, we have
x l3 4, the largest integer 3, s2 1, 3 S O L U T I O N The graph of the greatest integer function is shown in Figure 6. Since x y=[ x ] lim x x l3 lim 3 3 Since x 2 for 2 x 3, we have
x l3 lim x x l3 lim 2 2
M FIGURE 6 Because these onesided limits are not equal, lim x l 3 x does not exist by Theorem 1. Greatest integer function The next two theorems give two additional properties of limits. Their proofs can be found in Appendix F.
2 THEOREM If f x t x when x is near a (except possibly at a) and the limits of f and t both exist as x approaches a, then
xla lim f x xla lim t x 3 THE SQUEEZE THEOREM If f x possibly at a) and
y tx h x when x is near a (except L h g xla lim f x xla lim h x L L then xla lim t x f
0 a x FIGURE 7 The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinching Theorem, is illustrated by Figure 7. It says that if t x is squeezed between f x and h x near a, and if f and h have the same limit L at a, then t is forced to have the same limit L at a. 106  C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S V EXAMPLE 11 Show that lim x 2 sin
xl0 1 x 0. S O L U T I O N First note that we cannot use
 xl0 lim x 2 sin 1 x xl0 lim x 2 lim sin
xl0 1 x because lim x l 0 sin 1 x does not exist (see Example 4 in Section 2.2). However, since 1
y sin 1 x 1 y=≈ we have, as illustrated by Figure 8, x2 x 2 sin 1 x lim x2 0 x We know that
xl0 lim x 2 0 and xl0 x2 0 y=_ ≈ FIGURE 8 Taking f x obtain x 2, t x x 2 sin 1 x , and h x lim x 2 sin 1 x x 2 in the Squeeze Theorem, we 0
M y=≈ sin(1/x) xl0 2.3 EXERCISES
(c) lim f x t x 2
xl2 1. Given that
xl2 lim f x 4 xl2 lim t x lim h x 0 x l0 (d) lim xl 1 fx tx fx (e) lim x 3f x
x l2 3 ( f ) lim s3
x l1 ﬁnd the limits that exist. If the limit does not exist, explain why. (a) lim f x
xl2 5t x (b) lim t x
xl2 3 – 9 Evaluate the limit and justify each step by indicating the (c) lim sf x
xl2 (d) lim xl2 3f x tx txhx fx appropriate Limit Law(s).
3. lim 3x 4
xl 2 2x 2
3 sx ) 2 x 6x 2
3 1 x3 4. lim
x l2 2x 2 1 x 6x 4
2 (e) lim
x l2 tx hx ( f ) lim xl2 5. lim (1
xl8 6. lim t 2
tl 1 1 3 t 3 5 2. The graphs of f and t are given. Use them to evaluate each limit, if it exists. If the limit does not exist, explain why.
y y 7. lim
x l1 1 1 3x 4 x 2 3x 4 x2 8. lim su 4
ul 2 3u 6 y=ƒ 1 1 x y=©
0 1 1 x 9. lim s16
x l4 10. (a) What is wrong with the following equation? x2 (a) lim f x
x l2 x x 2 6 tx (b) lim f x
x l1 tx x 3 S E C T I O N 2 . 3 C A L C U L AT I N G L I M I T S U S I N G T H E L I M I T L AW S  107 (b) In view of part (a), explain why the equation lim
x l2 x2 x x 2 6 functions f x the same screen. x 2, t x x 2 cos 20 x, and h x x 2 on lim x
x l2 3 is correct.
11– 30 Evaluate the limit, if it exists. 11. lim
x l2 ; 34. Use the Squeeze Theorem to show that
lim sx 3
x l0 x 2 sin x 0 x2 x x
2 x 2 x x t2 2 6 6 9 7t 3 16 12. lim 14. lim xl 4 x2 x2 x2 x2 x2 x
2 5x 3x 4 4 Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen.
35. If 4 x 36. If 2 x 13. lim
x l2 xl4 4x 3x 4 4x 3x 4 9 tx fx x4 x2 x2 2 x
x 4x 7 for x 0, ﬁnd lim x l 4 f x . 2 for all x, evaluate lim x l 1 t x . 0. 0. 15. lim 17. lim tl 3 2t 4 2 16. lim 18. lim
x l1 37. Prove that lim x 4 cos
x l0 xl 1 hl0 h2 h 2 8 x3 x2 2 s1 x2 1 1 h3 h h h 2x x4 1 1 t2 h h 5
1 38. Prove that lim sx e sin
x l0 39 – 44 Find the limit, if it exists. If the limit does not exist, x 19. lim 3 xl 2 x 20. lim 8 1 1 explain why.
39. lim (2 x
xl3 h l0 9t 21. lim tl9 3 st
23. lim
x l7 x 2x 2x 3 3 1 x2 1 x ) 40. lim 22. lim xl 6 2x x 2 2 1 x 12 6 x x 1 x h l0 sx 2 x 7 1 x x 3 24. lim 41. lim
x l 0.5 42. lim xl 2 xl 1 1 4 25. lim xl 4 4
27. lim 29. lim
tl0 26. lim
tl0 1 t 3 43. lim
x l0 1 x 44. lim
x l0 t 3
1 x l 16 4 sx 16 x x 2 1 t s1 t 1 t 28. lim 45. The signum (or sign) function, denoted by sgn, is deﬁned by hl0 sx 2 9 30. lim xl 4 x4 sgn x 1 0 1 if x if x if x 0 0 0 ; 31. (a) Estimate the value of
lim
x l0 x (s1 3x 1). by graphing the function f x (b) Make a table of values of f x for x close to 0 and guess the value of the limit. (c) Use the Limit Laws to prove that your guess is correct. s1 x 3x 1 (a) Sketch the graph of this function. (b) Find each of the following limits or explain why it does not exist. (i) lim sgn x (ii) lim sgn x
x l0 x l0 (iii) lim sgn x
xl0 (iv) lim sgn x
xl0 46. Let ; 32. (a) Use a graph of
fx fx s3 x x s3 4 x x2 1 if x if x 2 2 to estimate the value of lim x l 0 f x to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Use the Limit Laws to ﬁnd the exact value of the limit. (a) Find lim x l 2 f x and lim x l 2 f x . (b) Does lim x l 2 f x exist? (c) Sketch the graph of f .
47. Let F x x2 x 1 . 1 (ii) lim F x
x l1 (a) Find (i) lim F x
x l1 ; 33. Use the Squeeze Theorem to show that
lim x l 0 x 2 cos 20 x 0. Illustrate by graphing the 108  C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S (b) Does lim x l 1 F x exist? (c) Sketch the graph of F.
48. Let 55. If lim xl1 fx x fx x2 8 1 10, ﬁnd lim f x .
xl1 56. If lim xl0 5, ﬁnd the following limits. (b) lim
xl0 tx x 3 2 x x2 3 if if if if x x 1 x 1 1 x 2 (a) lim f x
xl0 fx x 2 57. If (a) Evaluate each of the following limits, if it exists. (i) lim t x (ii) lim t x (iii) t 1
xl1 xl2 xl1 xl2 fx prove that lim x l 0 f x x2 0 0. if x is rational if x is irrational (iv) lim t x (b) Sketch the graph of t.
49. (a) If the symbol
xl 2 (v) lim t x (vi) lim t x
xl2 58. Show by means of an example that lim x l a f x denotes the greatest integer function deﬁned in Example 10, evaluate (i) lim x (ii) lim x (iii) lim x
xl 2 x l 2.4 t x may exist even though neither limx l a f x nor limx l a t x exists. exist even though neither lim x l a f x nor limx l a t x exists.
xl2 59. Show by means of an example that limx l a f x t x may (b) If n is an integer, evaluate (i) lim x (ii) lim x
x ln xln 60. Evaluate lim (c) For what values of a does lim x l a x exist?
50. Let f x s6 s3 x x 2 . 1 cos x , x . (a) Sketch the graph of f. (b) Evaluate each limit, if it exists. (i) lim f x (ii) lim
xl0 xl 61. Is there a number a such that xl 2 lim 3x 2 x2 ax a 3 x2 2 fx exists? If so, ﬁnd the value of a and the value of the limit.
62. The ﬁgure shows a ﬁxed circle C1 with equation (iii) xl lim 2 fx (iv) lim f x
xl 2 (c) For what values of a does lim x l a f x exist?
51. If f x x equal to f 2 . x , show that lim x l 2 f x exists but is not 52. In the theory of relativity, the Lorentz contraction formula x 1 2 y 2 1 and a shrinking circle C2 with radius r and center the origin. P is the point 0, r , Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the xaxis. What happens to R as C2 shrinks, that is, as r l 0 ?
y L L 0 s1 v2 c2 expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of the object at rest and c is the speed of light. Find lim v l c L and interpret the result. Why is a lefthand limit necessary?
53. If p is a polynomial, show that lim xl a p x P C™
0 Q pa. C¡ R x 54. If r is a rational function, use Exercise 53 to show that lim x l a r x r a for every number a in the domain of r. ...
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This note was uploaded on 12/16/2009 for the course MATH 1014 taught by Professor Ganong during the Spring '09 term at York University.
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