If x is within 0.0019 of 2, then 8 x is within

9. If x is within 0.0019 of 2, then 8 x is within

59. lim x ; The computer gives a value of 0, but

0.002 of 4.

lim x does not exist.

1.2 Concepts Review

1. L – ε ; L+ ε

2. 0<x–a< δ ; f ( x) – L < ε

is within 0.002 ε

10. If x is within 0.001 of 2, then

3. of 4.

4. ma + b

Problem Set 1.2

1. 0 < t – a < δ ⇒ f ( t ) – M < ε

2. 0 < u – b <⇒ δ gu ()– L < ε 11. 0 < x –0 <⇒ δ (2 – 1) – (–1) x < ε 2–11 x +<⇔ ε 2 x < ε

(2 – 1) – (–1) x = 2 x = 2 x < 2 δε =

68 Section 1.2 Instructor’s Resource Manual

12. 0 <+ x 21 <⇒ δ (3 – 1) – (–64) x < ε 2 x 2 – 11 x + 5

(3 – 1) – (–64) x = 3 x + 63 = 3 x + 21 < 3 δε =

= 2–1–9 x = 2( – 5) x = 2 x –5 < 2 δε =

⇔ 2–11 x +< ε 2 x + 2

( 2 – 1 – 7( – 3))( 2 – 1 x x x + 7( – 3)) x

Instructor’s Resource Manual

Section 1.2 69

To bound

, agree that

x – 3( 2 – 1 x + 7( – 3)) x

( –1) x 7 9 (10 – 6)( – 1) x x 2

δ ≤ 1 . If δ ≤ 1 , then <x< , so

hence x −⋅ 4 < ε ⇔ 10 x –1 < ε

1.65 For whatever ε is chosen, let δ be the smaller of

only when ≤

2 To bound 2 x + 2 , agree that δ ≤ 14 . x – 20 x + 6 1

x –1 < δ implies

2 x += 2 2–24 x +

70 Section 1.2 Instructor’s Resource Manual

21. <+<⇒ x δ 0 2 1 ( x – 2 – 1) – 2 x < ε 2 ⎛⎞ 25. For all 1 x ≠, 0 ≤ 0 sin

⎜⎟ ≤ 1 so

⎝⎠ x

⎜⎟ 4 ⎛⎞≤ x for all x ≠ . By Problem 18, 0

x 2 –2–1–2 x = x 2 –2–3 x =+ x 1 x –3 4 2 1

x sin

To bound x –3 , agree that δ ≤ 1 .

⎝⎠ x

x +< 1 δ implies

lim x = 0, so, by Problem 20,

x →0

x –3 =+ x 1–4 ≤++ x 1 –4 <1+4=5

2 2 26. 0 < x < δ ⇒ x – 0 = x = x < ε ( x – 2 – 1) – 2 x = x –2–3 x

x 4 = xx 3 . To bound x 3 , agree that

27. lim + x 0 : < x < δ ⇒ x – 0 < ε

1. x < δ ≤ 1 implies x 3 = x 3 ≤

δ = min{1, }; 0 ε < x <⇒ δ x 4 = xx 3 <⋅ ε 1 Thus, lim x →0 + x = 0.

lim x :0 < 0– x <⇒ δ x –0 < ε

23. Choose ε > 0. Then since lim ( ) fx = L , there is

x = x x → c For x < 0, x = –x; note also that

some δ 1 > 0 such that

since x ≥ 0.

0 < x – c < δ 1 ⇒ fx ()– L < ε .

δε = ;0 <−<⇒ x δ x = x =−<= x δε

Since lim f (x) = M, there is some δ 2 > 0 such

Thus, lim

that 0 <−< x c δ 2 ⇒ fx () − M < ε .

since lim x = lim x =

0, lim x = 0.

Let δ = min{δ 1 ,δ 2 } and choose x 0 such that

0<x 0 –c< δ.

28. Choose ε > 0. Since lim g ( x) = 0 there is some

Thus, fx ( 0 )– L <⇒−< ε ε fx ( 0 ) −< L ε x →a δ 1 >0 such that

⇒− fx ( 0 ) −<−<− ε L fx ( 0 ) + ε

0<x–a<δ ⇒ g(x ) − 0 < ε .

⇒ fx ( 0 ) −<< ε L fx ( 0 ) + ε

Let δ = min{1, δ 1 } , then fx () < B for

Similarly,

fx ( 0 ) −< ε M < fx ( 0 ) + ε .

x −< a δ or x −<⇒ a δ fx () < B . Thus,

Thus,

x −<⇒ a δ fxgx ()()0 −= fxgx ()()

− 2 ε <− L M < 2. ε As ε ⇒ 0, L − M → 0, so

= fx () gx () <⋅= B εε so lim f ( x)g(x) = 0. B x →a

L = M.

24. Since lim G (x) = 0, then given any ε > 0, we

x →c

can find δ > 0 such that whenever x – c < δ ,() Gx < ε .

Take any ε > 0 and the corresponding δ that

works for G(x), then x – c < δ implies

Fx ()–0 = Fx () ≤ Gx () < ε since

lim G (x) = 0.

x →c

Thus, lim F ( x) = 0.

x →c

Instructor’s Resource Manual

Section 1.2 71

29. Choose ε > 0. Since lim f ( x) = L, there is a

1.3 Concepts Review

x →a

δ > 0 such that for 0 < x – a < δ, fx ()– L < ε . 1. 48

That is, for

Let f(a) = A,

M = max L − ε , L + ε , A , c=a– δ,

d =a+ δ. Then for x in (c, d), fx () ≤ M , since either x = a, in which case

Problem Set 1.3

30. Suppose that L > M. Then L – M = α > 0. Now

and δ = min{ δ 1 , δ 2 } where

Thus, for 0 < x – a < δ , x → –1

x →–1

L– = 3 lim x – lim 1 ε < f(x) < L + ε and M – ε < g(x) < M + ε. 8

x → –1

x →–1

Combine the inequalities and use the fact

that fx () ≤ gx () to get

L – ε < f(x) ≤ g(x) < M + ε which leads to

L – ε < M + ε or L – M < 2ε. However,

6 L –M= α > 2ε

3. lim [(2 x +1)( x – 3)]

x →0

which is a contradiction. = lim (2 x +1) ⋅ lim (x – 3) 4, 5 Thus L ≤ M .

31. (b) and (c) are equivalent to the definition of

32. For every ε > 0 and δ > 0 there is some x with

0 < x – c < δ such that f (x ) – L > ε .

2 4. lim [(2 x 1)(7 x +

33. a. g (x) =

–x –2x–4

b. No, because

an asymptote at x ≈ 3.49.

= ⎢ 2 ⎜ lim x ⎟ + 1 ⎥ ⎢ 7 ⎜ lim x ⎟ + 13 ⎥

c. If δ≤ , then 2.75 < x < 3

4 or 3 < x < 3.25 and by graphing

on the interval [2.75, 3.25], we see that

3 x 2 – x –2–4 x 0 <

x –4 3 x 2 + x ++ x 6

so m must be at least three.