The Precise Definition of a Limit

EXAMPLE 5 Finding Delta Algebraically Prove that lim x:2 ƒsxd = 4 if

ƒsxd = e

Our task is to show that given P70 there exists a d 70 such that for all x

y ⫽x

06 ƒx-2ƒ6d

ƒ ƒsxd - 4 ƒ 6 P .

1. Solve the inequality ƒ ƒsxd - 4 ƒ 6 P to find an open interval containing x 0 =2 on

which the inequality holds for all xZx 0 .

For xZx 0 = 2, we have ƒsxd = x 2 , and the inequality to solve is ƒx 2 - 4 ƒ 6 P:

ƒx 2 -4ƒ6P

-P 6 x 2 -46P

4-P6x 2 64+P

Assumes see P64; below. 兹4 ⫺ ⑀

24 - P 6 ƒ x ƒ 6 24 + P

An open interval about x 0 =2 that solves the inequality FIGURE 2.20 An interval containing

24 - P 6 x 6 24 + P .

The inequality ƒ ƒsxd - 4 ƒ 6 P holds for all xZ 2 in the open interval A 24 - P,

x= 2 so that the function in Example 5

satisfies ƒ ƒsxd - 4 ƒ 6 P .

24 + P B (Figure 2.20).

2. Find a value of d 70 that places the centered interval s2 - d, 2 + dd inside the in-

terval A 24 - P, 24 + P B .

Take to be the distance from d x 0 =2 to the nearer endpoint of A 24 - P, 24 + P B . In other words, take d = min E 2 - 24 - P, 24 + P - 2 F , the minimum (the smaller)

of the two numbers 2 - 24 - P and

24 + P - 2. If has this or any smaller positive d value, the inequality 06 ƒx-2ƒ6d will automatically place x between and 24 - P

24 + P to make ƒ ƒsxd - 4 ƒ 6 P . For all x,

06 ƒx-2ƒ6d

ƒ ƒsxd - 4 ƒ 6 P .

This completes the proof.

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2.3 The Precise Definition of a Limit

Why was it all right to assume P 6 4? Because, in finding a such that for all d x ,06 ƒx-2ƒ6d implied ƒ ƒsxd - 4 ƒ 6 P 6 4, we found a that would work for d

any larger as well. P

Finally, notice the freedom we gained in letting d = min E 2 - 24 - P,

24 + P - 2 F . We did not have to spend time deciding which, if either, number was the

smaller of the two. We just let represent the smaller and went on to finish the argument. d

Using the Definition to Prove Theorems

We do not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples. Rather we appeal to general theorems about limits, in particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 2). As an example, we prove part 1 of Theorem 1, the Sum Rule.

EXAMPLE 6 Proving the Rule for the Limit of a Sum Given that lim x:c ƒsxd = L and lim x:c g sxd = M , prove that

lim sƒsxd + g sxdd = L + M .

x :c

Solution

Let P70 be given. We want to find a positive number such that for all x d

06 ƒx-cƒ6d

ƒ ƒsxd + gsxd - sL + Md ƒ 6 P .

Regrouping terms, we get

ƒ ƒsxd + gsxd - sL + Md ƒ = ƒ sƒsxd - Ld + sgsxd - Md ƒ

ƒ ƒsxd - L ƒ + ƒ gsxd - M ƒ. Triangle Inequality:

ƒa+bƒ…ƒaƒ+ƒbƒ

Since lim x:c ƒsxd = L , there exists a number d 1 70 such that for all x

06 ƒx-cƒ6d 1

ƒ ƒsxd - L ƒ 6 P

Similarly, since lim x:c g sxd = M , there exists a number d 2 70 such that for all x

06 ƒx-cƒ6d 2

ƒ gsxd - M ƒ 6 P

Let d = min 5d 1 ,d 2 6, the smaller of d 1 and d 2 . If 06 ƒx-cƒ6d then ƒx-cƒ6d 1 , so and ƒ ƒsxd - L ƒ 6 P >2, ƒx-cƒ6d 2 , so Therefore ƒ gsxd - M ƒ 6 P >2.

ƒ ƒsxd + gsxd - sL + Md ƒ 6 + 2 2 =P. This shows that lim x:c sƒsxd + g sxdd = L + M .

Let’s also prove Theorem 5 of Section 2.2. EXAMPLE 7 Given that lim x:c ƒsxd = L and lim x:c g sxd = M , and that ƒsxd … g sxd

for all x in an open interval containing c (except possibly c itself), prove that L…M .

Solution

We use the method of proof by contradiction. Suppose, on the contrary, that L7M . Then by the limit of a difference property in Theorem 1,

lim s g sxd - ƒsxdd = M - L .

x :c

Therefore, for any P 7 0, there exists d 70 such that

ƒ sgsxd - ƒsxdd - sM - Ld ƒ 6 P

whenever

0 6 ƒ x - c ƒ 6 d.

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98 Chapter 2: Limits and Continuity

Since L-M7 0 by hypothesis, we take P=L-M in particular and we have a number

d 70 such that ƒ sgsxd - ƒsxdd - sM - Ld ƒ 6 L - M

whenever

0 6 ƒ x - c ƒ 6 d.

Since a… ƒaƒ for any number a, we have

sg sxd - ƒsxdd - sM - Ld 6 L - M

whenever 06ƒx-cƒ6d

which simplifies to

g sxd 6 ƒsxd

whenever

0 6 ƒ x - c ƒ 6 d.

But this contradicts ƒsxd … g sxd . Thus the inequality L7M must be false. Therefore L…M .

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98 Chapter 2: Limits and Continuity

EXERCISES 2.3

Centering Intervals About a Point

In Exercises 1–6, sketch the interval (a, b) on the x-axis with the

point x 0 inside. Then find a value of d such that for all

⑀ 1 L ⫽ y ⫽ 兹x ⑀ ⫽4 1. a= 1,

x , 06 ƒx-x 0 ƒ6d Q a6x6b .

3. a=- 7 >2, b = -1>2, x 0 =-3

4. a=- 7 >2, b = -1>2, x 0 =-3 >2 5. a= 4 >9, b = 4>7, x 0 =1 >2

a= 2.7591,

b = 3.2391, x 0 =3

Finding Deltas Graphically x –1 0 2.61 3 3.41

Q ƒ ƒsxd - L ƒ 6 P .

In Exercises 7–14, use the graphs to find a d 70 such that for all x

NOT TO SCALE

06 ƒx-x 0 ƒ6d

f (x)

2 f ⫽x (x) ⫽4⫺x f (x) ⫽– x⫹3 2 x 0 ⫽2 x 0 ⫽ –1

0 ⫽5 3 y ⫽x 2 y ⫽4⫺x 5.8 3 L ⫽6 y 5 ⫽– x⫹3

NOT TO SCALE

NOT TO SCALE

NOT TO SCALE

NOT TO SCALE

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2.3 The Precise Definition of a Limit

13. 14. More on Formal Limits

Each of Exercises 31–36 gives a function ƒ(x), a point , and a posi- x 0 2

70 兹–x

tive number P. Find L= lim ƒsxd . Then find a number d such

that for all x

0 ⫽ –1 L ⫽2

ƒ ƒsxd - L ƒ 6 P . ⑀

06 ƒx-x ƒ6d

0 2 = 3, P = 0.02

31. ƒsxd = 3 - 2 x,

P = 0.03 兹–x

2 32. ƒsxd = - 3x - 2,

x 0 = - 1,

0 = 2, P = 0.05

1.99 ƒsxd = 33. x x- -4 2 ,

x+ 5 , x 0 = - 5, P = 0.05

x + 6x + 5

34. ƒsxd =

0 = - 3, P = 0.5

35. ƒsxd = 21 - 5x,

0 = 2, P = 0.4

36. ƒsxd = 4 >x,

Prove the limit statements in Exercises 37–50.

0 1 37. lim s9 - xd = 5

38. lim s3x - 7d = 2 – 16 –1 – 16 0 1 2 1 x :4 x :3

9 25 2.01 1.99 39. lim 2x - 5 = 2

40. lim 24 - x = 2

NOT TO SCALE

x :9

x :0

41. lim ƒsxd = 1 if ƒsxd =

Finding Deltas Algebraically

42. lim ƒsxd = 4 if ƒsxd =

e 1,

xZ- 2

Each of Exercises 15–30 gives a function ƒ(x) and numbers L ,x 0 and

x : -2

x=- 2

P 7 0. In each case, find an open interval about x 0 on which the in-

equality ƒ ƒsxd - L ƒ 6 P holds. Then give a value for d 70 such

43. lim 1 x=1

that for all x satisfying the 06 ƒx-x 0 ƒ6d inequality x :1

ƒ ƒsxd - L ƒ 6 P holds.

0 = 4, P = 0.01

44. 1 lim = 1

15. ƒsxd = x + 1, L= 5,

x : 23 x 2 3

0 = - 2, P = 0.02 45. x : -3 lim x+ 3 =-6 46. lim x- 1 =2

16. ƒsxd = 2x - 2, L=- 6,

0 = 0, P = 0.1

x 2 -9

4 - 2x, x6 1

x 2 -1

17. ƒsxd = 2x + 1, L= 1,

>2, 0 =1 >4, P = 0.1 47. x :1 if ƒsxd = e 6x - 4, xÚ 1

x :1

18. ƒsxd = 2x,

L= 1 x

0 = 10, P=1

lim ƒsxd = 2

19. ƒsxd = 219 - x,

L= 3,

0 = 23, P=1

x6 0 20. ƒsxd = 2x - 7,

48. lim ƒsxd = 0 if ƒsxd = e

2x,

xÚ 0 21. ƒsxd = 1 >x,

L= 4,

0 = 4, 2 P = 0.05 49. lim x sin 1

22. ƒsxd = x , L= 3,

x 0 = 23,

0 = - 2, P = 0.5

23. ƒsxd = x 2

, L= 4,

0 = - 1, P = 0.1

24. ƒsxd = 1 >x, L=- 1,

0 = 4, P=1

25. ƒsxd = x 2

- 5, L= 11,

0 = 24, P=1

26. ƒsxd = 120 >x, L= 5,

y 1 27. ƒsxd = mx,

m7 0,

L= 2m,

28. ƒsxd = mx, m7 0,

L= 3m,

P=c70

␲ 29. ƒsxd = mx + b,

x >2d + b,

0 =1 >2, P=c70

m7 0,

L = sm

30. ƒsxd = mx + b, m7 0,

L=m+b ,

P = 0.05

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Chapter 2: Limits and Continuity

1 volts and I is to be 5 ; 0.1 amp . 50. In what interval does R have to lim x 2 sin x=0 x :0

lie for I to be within 0.1 amp of the value I 0 =5?

1 y ⫽x 2 V I R

When Is a Number L Not the Limit of ƒ(x)

y ⫽x 2 sin 1 x

as x:x 0 ?

– 2 0 2 1 We can prove that lim x:x 0 ƒsxd Z L by providing an P70 ␲ such that ␲ no possible d 70 satisfies the condition

Q ƒ ƒsxd - L ƒ 6 P .

For all x, 06ƒx-x 0 ƒ6d

We accomplish this for our candidate P by showing that for each

d 70 there exists a value of x such that

y ⫽ –x ƒ ƒsxd - L ƒ Ú P .

2 06 ƒx-x 0 ƒ6d

and

y ⫽ f(x)

Theory and Examples

51. Define what it means to say that lim

g sxd = k .

L ⫹⑀

52. Prove that x :0 lim ƒsxd = L if and only if lim ƒsh + cd = L .

53. A wrong statement about limits Show by example that the fol- lowing statement is wrong.

The number L is the limit of ƒ(x) as x approaches if x 0 ƒ(x) gets

f (x)

closer to L as x approaches x 0 .

Explain why the function in your example does not have the given x

0 x 0 ⫺␦ x 0 x 0 ⫹␦ 54. Another wrong statement about limits Show by example that

value of L as a limit as x:x 0 .

a value of x for which the following statement is wrong.

0 ⬍ 兩 x ⫺x 0 兩 ⬍ ␦ and 兩 f (x) ⫺L 兩 ⱖ ⑀

The number L is the limit of ƒ(x) as x approaches x 0 if, given any

P 7 0, there exists a value of x for which ƒ ƒsxd - L ƒ 6 P .

e x+ 1, x7 1.

x6 1

57. Let ƒsxd =

Explain why the function in your example does not have the given

value of L as a limit as x:x 0 .

55. Grinding engine cylinders Before contracting to grind engine cylinders to a cross-sectional area of 9 in 2

y ⫽x⫹1

, you need to know how

much deviation from the ideal cylinder diameter of x 0 = 3.385

in. you can allow and still have the area come within

0.01 in 2 of

the required 9 in 2 . To find out, you let

A = psx 2 >2d 2 and look for

y ⫽ f(x) the interval in which you must hold x to make ƒ A - 9 ƒ … 0.01 . What interval do you find? 1

56. Manufacturing electrical resistors Ohm’s law for electrical cir- x

cuits like the one shown in the accompanying figure states that

V = RI . In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your firm has been

y ⫽x

asked to supply the resistors for a circuit in which V will be 120

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2.3 The Precise Definition of a Limit

60. a. For the function graphed here, show that lim x : -1 g sxd Z 2 . following condition:

a. Let P=1 >2. Show that no possible d 70 satisfies the

For all x, 06ƒx-1ƒ6d Q ƒ ƒsxd - 2 ƒ 6 1 >2. the limit? If not, why not?

b. Does lim x : -1 g sxd appear to exist? If so, what is the value of

That is, for each d 70 show that there is a value of x such that

06 ƒx-1ƒ6d and ƒ ƒsxd - 2 ƒ Ú 1 >2.

This will show that lim x:1 ƒsxd Z 2 .

b. Show that lim x:1 ƒsxd Z 1 . c. Show that lim x:1 ƒsxd Z 1.5 .

y ⫽ g(x)

x , x6 2

58. Let hsxd = • 3, x= 2 2, x7 2.

4 y ⫽ h(x)

COMPUTER EXPLORATIONS

3 In Exercises 61–66, you will further explore finding deltas graphi- y ⫽2 cally. Use a CAS to perform the following steps:

2 a. Plot the function y= ƒsxd near the point x 0 being approached. 1 y ⫽x 2 b. Guess the value of the limit L and then evaluate the limit

symbolically to see if you guessed correctly.

0 2 c. Using the value P = 0.2 , graph the banding lines y 1 =L-P and y 2 =L+P together with the function ƒ near x 0 .

Show that d. From your graph in part (c), estimate a d 70 such that for all x a. lim hsxd Z 4

ƒx-x 0 ƒ6d Q ƒ ƒsxd - L ƒ 6 P .

x :2

b. lim hsxd Z 3

Test your estimate by plotting ƒ, y 1 , and y 2 over the interval c. lim hsxd Z 2

x :2

06 ƒx-x 0 ƒ6d. For your viewing window use

x 0 - 2d … x … x 0 + 2d and If L- 2P … y … L + 2P . any 59. For the function graphed here, explain why

x :2

function values lie outside the interval [L - P, L + P] , your a. lim ƒsxd Z 4

choice of was too large. Try again with a smaller estimate. d

x :3

b. lim ƒsxd Z 4.8 e. Repeat parts (c) and (d) successively for P = 0.1, 0.05 , and

x :3

c. lim ƒsxd Z 3

ƒsxd = 5x 3 + 9x 62. 2 5 2 , x 0 =0 2x + 3x

y ⫽ f(x)

xs 1 - cos xd

64. ƒsxd = x- sin , x x 0 =0

2 3 x- 1 65. ƒsxd = x- 1 , x 0 =1

3x x 2 - s7x + 1d2x + 5

0 3 66. ƒsxd =

x- 1 , x 0 =1

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Chapter 2: Limits and Continuity