1. The document provides examples and concepts related to limits, including definitions, evaluating limits of various functions as the variable approaches a number, and evaluating one-sided and two-sided limits. 2. Worked examples evaluate limits of rational functions, radical functions, and trigonometric functions as the variable approaches numbers or infinity. 3. Key concepts such as left-hand and right-hand limits, and limits at infinity, are demonstrated through multiple examples.

1. CHAPTER 1 Limits
x3 – 4 x 2 + x + 6
1.1 Concepts Review 9. lim
x → –1 x +1
1. L; c ( x + 1)( x 2 – 5 x + 6)
= lim
x → –1 x +1
2. 6
= lim ( x 2 – 5 x + 6)
x → –1
3. L; right
= (–1) – 5(–1) + 6
2
4. lim f ( x) = M = 12
x →c
x 4 + 2 x3 – x 2
10. lim
Problem Set 1.1 x →0 x2
= lim( x 2 + 2 x –1) = –1
1. lim( x – 5) = –2 x →0
x →3
2. lim (1 – 2t ) = 3 x2 – t 2 ( x + t )( x – t )
t → –1
11. lim = lim
x→–t x + t x→ – t x+t
= lim ( x – t )
3. lim ( x 2 + 2 x − 1) = (−2) 2 + 2(−2) − 1 = −1 x→ –t
x →−2 = –t – t = –2t
4. lim ( x 2 + 2t − 1) = (−2) 2 + 2t − 1 = 3 + 2t x2 – 9
x →−2 12. lim
x →3 x – 3
(
5. lim t 2 − 1 =
t →−1
) ( ( −1) − 1) = 0
2
= lim
x →3
( x – 3)( x + 3)
x–3
= lim( x + 3)
(
6. lim t 2 − x 2 =
t →−1
) ( ( −1) 2
)
− x2 = 1 − x2 x →3
=3+3=6
x2 – 4 ( x – 2)( x + 2) (t + 4)(t − 2) 4
7. lim = lim 13. lim
x→2 x – 2 x→2 x–2 t →2 (3t − 6) 2
= lim( x + 2)
x→2 (t − 2) 2 t + 4
= lim
=2+2=4 t →2 9(t − 2) 2
t 2 + 4t – 21 t+4
8. lim = lim
t →2 9
t → –7 t+7
(t + 7)(t – 3) 2+4 6
= lim = =
t → –7 t+7 9 9
= lim (t – 3)
t → –7
(t − 7)3
= –7 – 3 = –10 14. lim
t →7+ t −7
(t − 7) t − 7
= lim
t →7 + t −7
= lim t −7
t →7+
= 7−7 = 0
Instructor’s Resource Manual Section 1.1 63
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or by any means, without permission in writing from the publisher.

2. x 4 –18 x 2 + 81 ( x 2 – 9) 2 1 − cos t
15. lim = lim lim =0
x →3 2 x →3 2 t →0 2t
( x – 3) ( x – 3)
( x – 3) 2 ( x + 3) 2 ( x − sin x ) 2 / x 2
= lim 2
= lim( x + 3)2 = (3 + 3) 2 21. x
x →3 ( x – 3) x →3
1. 0.0251314
= 36
0.1 2.775 × 10−6
(3u + 4)(2u – 2)3 8(3u + 4)(u –1)3 0.01 2.77775 × 10−10
16. lim = lim
u →1 (u –1) 2 u →1 (u –1) 2 0.001 2.77778 × 10−14
= lim 8(3u + 4)(u – 1) = 8[3(1) + 4](1 – 1) = 0
u →1 –1. 0.0251314
–0.1 2.775 × 10−6
(2 + h) 2 − 4 4 + 4h + h 2 − 4
17. lim = lim –0.01 2.77775 × 10−10
h→0 h h→0 h
h 2 + 4h –0.001 2.77778 × 10−14
= lim = lim(h + 4) = 4
h →0 h h →0 ( x – sin x) 2
lim =0
x →0 x2
( x + h) 2 − x 2 x 2 + 2 xh + h 2 − x 2
18. lim = lim 2 2
h→0 h h →0 h 22. x (1 − cos x ) / x
h 2 + 2 xh 1. 0.211322
= lim = lim(h + 2 x) = 2 x
h →0 h h →0 0.1 0.00249584
0.01 0.0000249996
19. x sin x 0.001 2.5 × 10−7
2x
1. 0.420735
–1. 0.211322
0.1 0.499167
–0.1 0.00249584
0.01 0.499992 –0.01 0.0000249996
0.001 0.49999992 –0.001 2.5 × 10−7
(1 – cos x) 2
–1. 0.420735 lim =0
x →0 x2
–0.1 0.499167
2
23. t (t − 1) /(sin(t − 1))
–0.01 0.499992
–0.001 0.49999992 2. 3.56519
1.1 2.1035
sin x
lim = 0.5 1.01
x →0 2 x 2.01003
1.001 2.001
1− cos t
20. t 2t
1. 0.229849 0 1.1884
0.1 0.0249792 0.9 1.90317
0.01 0.00249998 0.99 1.99003
0.999 1.999
0.001 0.00024999998
t −12
lim =2
–1. –0.229849 t →1 sin(t − 1)
–0.1 –0.0249792
–0.01 –0.00249998
–0.001 –0.00024999998
64 Section 1.1 Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.

3. 27. x ( x − π / 4) 2 /(tan x − 1) 2
x −sin( x − 3) − 3
24. x x −3
1. + π
4
0.0320244
π
4. 0.158529 0.1 + 4 0.201002
π
3.1 0.00166583 0.01 + 4 0.245009
π
3.01 0.0000166666 0.001 + 4 0.2495
3.001 1.66667 × 10−7
−1. + π
4 0.674117
2. 0.158529 −0.1 + π
4 0.300668
2.9 0.00166583 −0.01 + π
4 0.255008
2.99 0.0000166666 −0.001 + π
4 0.2505
2.999 1.66667 × 10−7
lim
(x − ) π 2
4
= 0.25
lim
x – sin( x – 3) – 3
=0 x→ π
4 (tan x − 1)2
x →3 x–3
28. u (2 − 2sin u ) / 3u
25. x (1 + sin( x − 3π / 2)) /( x − π )
1. + π 0.11921
1. + π 0.4597 2
0.1 + π 0.00199339
0.1 + π 0.0500
2
0.01 + π 0.0000210862
0.01 + π
2
0.0050 0.001 + π
2.12072 × 10−7
2
0.001 + π 0.0005
−1. + π
2 0.536908
–1. + π –0.4597 −0.1 + π
2 0.00226446
–0.1 + π –0.0500 −0.01 + π
0.0000213564
2
–0.01 + π –0.0050 −0.001 + π
2.12342 × 10−7
2
–0.001 + π –0.0005 2 − 2sin u
lim =0
1 + sin ( x − 32π ) u→ π 3u
2
lim =0
x →π x−π
29. a. lim f ( x) = 2
x → –3
26. t (1 − cot t ) /(1 / t )
b. f(–3) = 1
1. 0.357907
0.1 –0.896664 c. f(–1) does not exist.
0.01 –0.989967
5
0.001 –0.999 d. lim f ( x) =
x → –1 2
–1. –1.64209 e. f(1) = 2
–0.1 –1.09666 f. lim f(x) does not exist.
x→1
–0.01 –1.00997
–0.001 –1.001 g. lim f ( x) = 2
x →1–
1 – cot t
lim 1
= –1
t →0
t h. lim f ( x) = 1
x →1+
5
i. lim f ( x ) =
+ 2
x →−1
Instructor’s Resource Manual Section 1.1 65
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or by any means, without permission in writing from the publisher.

4. 30. a. lim f ( x) does not exist. b. lim f ( x) does not exist.
x → –3 x →1
b. f(–3) = 1 c. f(1) = 2
c. f(–1) = 1 d. lim f ( x) = 2
x →1+
d. lim f ( x) = 2
x → –1
34.
e. f(1) = 1
f. lim f ( x) does not exist.
x →1
g. lim f ( x) = 1
x →1–
h. lim f ( x) does not exist.
x →1+
i. lim f ( x ) = 2
x →−1+
a. lim g ( x) = 0
31. a. f(–3) = 2 x →1
b. f(3) is undefined. b. g(1) does not exist.
c. lim f ( x) = 2
x → –3− c. lim g ( x ) = 1
x→2
d. lim f ( x) = 4
x → –3+ d. lim g ( x ) = 1
x → 2+
e. lim f ( x) does not exist.
x → –3
35. f ( x) = x – ⎡[ x ]⎤
⎣ ⎦
f. lim f ( x) does not exist.
x →3+
32. a. lim f ( x) = −2
x → –1−
b. lim f ( x) = −2
x → –1+
c. lim f ( x) = −2
x → –1
d. f (–1) = –2
e. lim f ( x) = 0
x →1 a. f(0) = 0
f. f (1) = 0 b. lim f ( x) does not exist.
x →0
33.
c. lim f ( x ) = 1
x →0 –
1
d. lim f ( x) =
x→ 1 2
2
a. lim f ( x) = 0
x →0
66 Section 1.1 Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.

5. x 41. lim f ( x) exists for a = –1, 0, 1.
36. f ( x) = x→a
x
42. The changed values will not change lim f ( x) at
x→a
any a. As x approaches a, the limit is still a 2 .
x −1
43. a. lim does not exist.
x →1 x −1
x −1 x −1
lim = −1 and lim =1
x →1 − x −1 +
x →1 x −1
x −1
b. lim = −1
a. f (0) does not exist. x →1 − x −1
b. lim f ( x) does not exist. x2 − x − 1 − 1
x →0 c. lim = −3
x →1− x −1
c. lim f ( x ) = –1
x →0 –
⎡ 1 1 ⎤
d. lim ⎢ − ⎥ does not exist.
− x −1 x −1 ⎥
d. lim f ( x) = 1 x →1 ⎢
⎣ ⎦
x→ 1
2
44. a. lim x− x =0
x2 − 1 x →1+
37. lim does not exist.
x →1 x − 1
1
x2 − 1 x2 − 1 b. lim does not exist.
lim = −2 and lim =2 x →0 + x
x →1− x −1 x →1+ x − 1
1/ x
x+2− 2 c. lim x(−1) =0
38. lim x → 0+
x →0 x
1/ x
( x + 2 − 2)( x + 2 + 2) d. lim x (−1) =0
= lim x →0+
x →0 x( x + 2 + 2)
x+2−2 x 45. a) 1 b) 0
= lim = lim
x →0 x( x + 2 + 2) x →0 x( x + 2 + 2)
c) −1 d) −1
1 1
2 1
= lim = = = 46. a) Does not exist b) 0
x →0 x+2+ 2 0+2 + 2 2 2 4
c) 1 d) 0.556
39. a. lim f ( x) does not exist.
x →1
47. lim x does not exist since x is not defined
x →0
b. lim f ( x) = 0
x →0 for x < 0.
40. 48. lim x x = 1
x → 0+
49. lim x =0
x →0
x
50. lim x = 1
x →0
sin 2 x 1
51. lim =
x →0 4 x 2
Instructor’s Resource Manual Section 1.2 67
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or by any means, without permission in writing from the publisher.

6. sin 5 x 5 7. If x is within 0.001 of 2, then 2x is within 0.002
52. lim = of 4.
x →0 3x 3
⎛1⎞
53. lim cos ⎜ ⎟ does not exist.
x →0 ⎝ x⎠
⎛1⎞
54. lim x cos ⎜ ⎟ = 0
x →0 ⎝ x⎠
x3 − 1
55. lim =6
x →1 2x + 2 − 2 8. If x is within 0.0005 of 2, then x2 is within 0.002
of 4.
x sin 2 x
56. lim =2
x →0 sin( x 2 )
x2 – x – 2
57. lim = –3
x →2– x–2
2
58. lim 1/( x −1)
=0
x →1 +
1+ 2
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.
x →0
lim x does not exist.
x →0−
1.2 Concepts Review
1. L – ε ; L + ε
2. 0 < x – a < δ ; f ( x) – L < ε 8
10. If x is within 0.001 of 2, then is within 0.002
ε x
3. of 4.
3
4. ma + b
Problem Set 1.2
1. 0 < t – a < δ ⇒ f (t ) – M < ε
2. 0 < u – b < δ ⇒ g (u ) – L < ε 11. 0 < x – 0 < δ ⇒ (2 x – 1) – (–1) < ε
2x – 1+ 1 < ε ⇔ 2x < ε
3. 0 < z – d < δ ⇒ h( z ) – P < ε ⇔ 2 x <ε
ε
4. 0 < y – e < δ ⇒ φ ( y ) – B < ε ⇔ x <
2
5. 0 < c – x < δ ⇒ f ( x) – L < ε
ε
δ = ;0 < x –0 <δ
2
6. 0 < t – a < δ ⇒ g (t ) – D < ε
(2 x – 1) – (–1) = 2 x = 2 x < 2δ = ε
68 Section 1.2 Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.

7. 12. 0 < x + 21 < δ ⇒ (3x – 1) – (–64) < ε 2 x 2 – 11x + 5
15. 0 < x – 5 < δ ⇒ –9 <ε
3 x – 1 + 64 < ε ⇔ 3 x + 63 < ε x–5
⇔ 3( x + 21) < ε 2 x 2 – 11x + 5 (2 x – 1)( x – 5)
–9 <ε ⇔ –9 <ε
⇔ 3 x + 21 < ε x–5 x–5
ε ⇔ 2x – 1 – 9 < ε
⇔ x + 21 <
3
⇔ 2( x – 5) < ε
ε ε
δ = ; 0 < x + 21 < δ ⇔ x–5 <
3 2
(3 x – 1) – (–64) = 3 x + 63 = 3 x + 21 < 3δ = ε
ε
δ = ;0 < x –5 <δ
2
x 2 – 25
13. 0 < x – 5 < δ ⇒ – 10 < ε 2 x – 11x + 5
2
(2 x – 1)( x – 5)
x–5 –9 = –9
x–5 x–5
x 2 – 25 ( x – 5)( x + 5) = 2 x – 1 – 9 = 2( x – 5) = 2 x – 5 < 2δ = ε
– 10 < ε ⇔ – 10 < ε
x–5 x–5
⇔ x + 5 – 10 < ε 16. 0 < x – 1 < δ ⇒ 2x – 2 < ε
⇔ x–5 <ε
2x – 2 < ε
δ = ε; 0 < x – 5 < δ ( 2 x – 2 )( 2 x + 2 )
⇔ <ε
2x + 2
2
x – 25 ( x – 5)( x + 5)
– 10 = – 10 = x + 5 – 10
x–5 x–5 2x – 2
⇔ <ε
= x–5 <δ =ε 2x + 2
x –1
⇔2 <ε
2x – x 2 2x + 2
14. 0 < x – 0 < δ ⇒ − (−1) < ε
x
2ε
2 x2 – x x(2 x – 1) δ= ; 0 < x –1 < δ
+1 < ε ⇔ +1 < ε 2
x x ( 2 x – 2)( 2 x + 2)
2x − 2 =
⇔ 2x – 1 +1 < ε 2x + 2
⇔ 2x < ε 2x – 2
=
⇔ 2 x <ε 2x + 2
ε 2 x –1 2 x – 1 2δ
⇔ x < ≤ < =ε
2 2x + 2 2 2
ε 2x – 1
δ = ;0 < x –0 <δ 17. 0 < x – 4 < δ ⇒ – 7 <ε
2 x–3
2 x2 – x x(2 x – 1)
− (−1) = + 1 = 2x – 1+ 1 2x – 1 2 x – 1 – 7( x – 3)
x x – 7 <ε ⇔ <ε
x–3 x–3
= 2 x = 2 x < 2δ = ε
( 2 x – 1 – 7( x – 3))( 2 x – 1 + 7( x – 3))
⇔ <ε
x – 3( 2 x – 1 + 7( x – 3))
2 x – 1 – (7 x – 21)
⇔ <ε
x – 3( 2 x – 1 + 7( x – 3))
–5( x – 4)
⇔ <ε
x – 3( 2 x – 1 + 7( x – 3))
Instructor’s Resource Manual Section 1.2 69
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or by any means, without permission in writing from the publisher.

8. ⇔ x−4 ⋅
5
<ε 10 x3 – 26 x 2 + 22 x – 6
19. 0 < x – 1 < δ ⇒ –4 <ε
x − 3( 2 x − 1 + 7( x − 3)) ( x – 1) 2
10 x3 – 26 x 2 + 22 x – 6
To bound
5
, agree that –4 <ε
x – 3( 2 x – 1 + 7( x – 3)) ( x –1)2
1 1 7 9 (10 x – 6)( x – 1)2
δ ≤ . If δ ≤ , then < x < , so ⇔ –4 <ε
2 2 2 2 ( x – 1)2
5 ⇔ 10 x – 6 – 4 < ε
0.65 < < 1.65 and
x – 3( 2 x – 1 + 7( x – 3))
⇔ 10( x – 1) < ε
5
hence x − 4 ⋅ <ε ⇔ 10 x – 1 < ε
x − 3( 2 x − 1 + 7( x − 3))
ε
ε ⇔ x –1 <
⇔ x–4 < 10
1.65
For whatever ε is chosen, let δ be the smaller of
ε
δ= ; 0 < x –1 < δ
1 ε 10
and .
2 1.65 10 x3 – 26 x 2 + 22 x – 6 (10 x – 6)( x – 1) 2
–4 = –4
⎧1 ε ⎫ ( x – 1) 2 ( x – 1) 2
δ = min ⎨ , ⎬, 0 < x – 4 < δ
⎩ 2 1. 65 ⎭ = 10 x − 6 − 4 = 10( x − 1)
2x −1 5 = 10 x − 1 < 10δ = ε
− 7 = x−4 ⋅
x −3 x − 3( 2 x − 1 + 7( x − 3))
< x – 4 (1.65) < 1. 65δ ≤ ε 20. 0 < x – 1 < δ ⇒ (2 x 2 + 1) – 3 < ε
1 1 ε
since δ = only when ≤ so 1.65δ ≤ ε .
2 2 1. 65 2 x2 + 1 – 3 = 2 x2 – 2 = 2 x + 1 x – 1
To bound 2 x + 2 , agree that δ ≤ 1 .
14 x 2 – 20 x + 6
18. 0 < x – 1 < δ ⇒ –8 < ε
x –1 x – 1 < δ implies
2x + 2 = 2x – 2 + 4
14 x 2 – 20 x + 6 2(7 x – 3)( x – 1)
–8 <ε ⇔ –8 <ε ≤ 2x – 2 + 4
x –1 x –1
<2+4=6
⇔ 2(7 x – 3) – 8 < ε
ε ⎧ ε⎫
⇔ 14( x – 1) < ε δ ≤ ; δ = min ⎨1, ⎬; 0 < x – 1 < δ
6 ⎩ 6⎭
⇔ 14 x – 1 < ε
(2 x + 1) – 3 = 2 x 2 – 2
2
ε
⇔ x –1 < ⎛ε ⎞
14 = 2x + 2 x −1 < 6 ⋅ ⎜ ⎟ = ε
⎝6⎠
ε
δ= ; 0 < x –1 < δ
14
14 x 2 – 20 x + 6 2(7 x – 3)( x – 1)
–8 = –8
x –1 x –1
= 2(7 x – 3) – 8
= 14( x – 1) = 14 x – 1 < 14δ = ε
70 Section 1.2 Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.

9. 21. 0 < x + 1 < δ ⇒ ( x 2 – 2 x – 1) – 2 < ε ⎛1⎞
25. For all x ≠ 0 , 0 ≤ sin 2 ⎜ ⎟ ≤ 1 so
⎝x⎠
x2 – 2 x – 1 – 2 = x2 – 2 x – 3 = x + 1 x – 3 ⎛ 1⎞
x 4 sin 2 ⎜ ⎟ ≤ x 4 for all x ≠ 0 . By Problem 18,
To bound x – 3 , agree that δ ≤ 1 . ⎝ x⎠
4
x + 1 < δ implies lim x = 0, so, by Problem 20,
x→0
x – 3 = x + 1 – 4 ≤ x + 1 + –4 < 1 + 4 = 5 4 2 ⎛ 1⎞
lim x sin ⎜ ⎟ = 0.
ε ⎧ ε⎫ x→0 ⎝ x⎠
δ ≤ ; δ = min ⎨1, ⎬ ; 0 < x + 1 < δ
5 ⎩ 5⎭
( x – 2 x – 1) – 2 = x 2 – 2 x – 3
2 26. 0 < x < δ ⇒ x –0 = x = x <ε
For x > 0, ( x ) = x.
2
ε
= x +1 x – 3 < 5⋅ =ε
5 x < ε ⇔ ( x )2 = x < ε 2
δ = ε 2; 0 < x < δ ⇒ x < δ = ε 2 = ε
22. 0 < x < δ ⇒ x 4 – 0 = x 4 < ε
x 4 = x x3 . To bound x3 , agree that 27. lim x : 0 < x < δ ⇒ x – 0 < ε
x →0 +
δ ≤ 1. x < δ ≤ 1 implies x3 = x ≤ 1 so
3 For x ≥ 0 , x = x .
δ = ε; 0 < x < δ ⇒ x – 0 = x = x < δ = ε
δ ≤ ε.
Thus, lim+ x = 0.
δ = min{1, ε }; 0 < x < δ ⇒ x 4 = x x3 < ε ⋅1 x→0
=ε lim x : 0 < 0 – x < δ ⇒ x – 0 < ε
x →0 –
23. Choose ε > 0. Then since lim f ( x) = L, there is For x < 0, x = – x; note also that x = x
x →c
some δ1 > 0 such that since x ≥ 0.
0 < x – c < δ1 ⇒ f ( x ) – L < ε . δ = ε ;0 < − x < δ ⇒ x = x = − x < δ = ε
Since lim f (x) = M, there is some δ 2 > 0 such Thus, lim– x = 0,
x→c x→0
that 0 < x − c < δ 2 ⇒ f ( x) − M < ε . since lim x = lim x = 0, lim x = 0.
Let δ = min{δ1 , δ2 } and choose x 0 such that x →0 + x →0 – x →0
0 < x0 – c < δ . 28. Choose ε > 0. Since lim g( x) = 0 there is some
Thus, f ( x0 ) – L < ε ⇒ −ε < f ( x0 ) − L < ε x→ a
δ1 > 0 such that
⇒ − f ( x0 ) − ε < − L < − f ( x0 ) + ε ε.
0 < x – a < δ1 ⇒ g(x ) − 0 <
⇒ f ( x0 ) − ε < L < f ( x0 ) + ε . B
Similarly, Let δ = min{1, δ1} , then f ( x) < B for
f ( x0 ) − ε < M < f ( x0 ) + ε . x − a < δ or x − a < δ ⇒ f ( x) < B. Thus,
Thus, x − a < δ ⇒ f ( x) g ( x) − 0 = f ( x) g ( x)
−2ε < L − M < 2ε . As ε ⇒ 0, L − M → 0, so
ε
L = M. = f ( x) g ( x) < B ⋅ = ε so lim f ( x)g(x) = 0.
B x→ a
24. Since lim G(x) = 0, then given any ε > 0, we
x→c
can find δ > 0 such that whenever
x – c < δ , G ( x) < ε .
Take any ε > 0 and the corresponding δ that
works for G(x), then x – c < δ implies
F ( x) – 0 = F ( x) ≤ G ( x ) < ε since
lim G(x) = 0.
x→c
Thus, lim F( x) = 0.
x→c
Instructor’s Resource Manual Section 1.2 71
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.

10. 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 , f ( x) – L < ε . 1. 48
That is, for
2. 4
a − δ < x < a or a < x < a + δ ,
L − ε < f ( x) < L + ε . 3. – 8; – 4 + 5c
Let f(a) = A,
M = max { L − ε , L + ε , A } , c = a – δ, 4. 0
d = a + δ. Then for x in (c, d), f ( x) ≤ M , since
either x = a, in which case
Problem Set 1.3
f ( x) = f (a ) = A ≤ M or 0 < x – a < δ so 1. lim (2 x + 1) 4
x→1
L − ε < f ( x) < L + ε and f ( x) < M .
= lim 2 x + lim 1 3
x→1 x→1
30. Suppose that L > M. Then L – M = α > 0. Now = 2 lim x + lim 1 2,1
α x →1 x→1
take ε < and δ = min{δ1 , δ 2} where = 2(1) + 1 = 3
2
0 < x – a < δ1 ⇒ f ( x) – L < ε and 2. lim (3x 2 – 1) 5
x→ –1
0 < x – a < δ 2 ⇒ g ( x) – M < ε .
= lim 3x 2 – lim 1 3
Thus, for 0 < x – a < δ , x→ –1 x→–1
= 3 lim x 2 – lim 1 8
L – ε < f(x) < L + ε and M – ε < g(x) < M + ε. x→ –1 x→–1
Combine the inequalities and use the fact ⎛ ⎞
2
that f ( x) ≤ g ( x) to get = 3⎜ lim x ⎟ – lim 1 2, 1
⎝ x→ –1 ⎠ x →–1
L – ε < f(x) ≤ g(x) < M + ε which leads to 2
= 3(–1) – 1 = 2
L – ε < M + ε or L – M < 2ε.
However, 3. lim [(2 x +1)( x – 3)] 6
L – M = α > 2ε x→0
which is a contradiction. = lim (2 x +1) ⋅ lim (x – 3) 4, 5
x→ 0 x→ 0
Thus L ≤ M .
⎛ ⎞ ⎛ ⎞
= ⎜ lim 2 x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟ 3
31. (b) and (c) are equivalent to the definition of ⎝ x→ 0 x→ 0 ⎠ ⎝ x→0 x→ 0 ⎠
limit. ⎛ ⎞ ⎛ ⎞
= ⎜ 2 lim x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟ 2, 1
⎝ x →0 x→ 0 ⎠ ⎝ x→0 x→ 0 ⎠
32. For every ε > 0 and δ > 0 there is some x with
= [2(0) +1](0 – 3) = –3
0 < x – c < δ such that f ( x ) – L > ε .
4. lim [(2 x 2 + 1)(7 x 2 + 13)] 6
x 3 – x 2 – 2x – 4 x→ 2
33. a. g(x) =
x 4 – 4x 3 + x 2 + x + 6 = lim (2 x 2 + 1) ⋅ lim (7 x 2 + 13) 4, 3
x→ 2 x→ 2
x+6 ⎛ ⎞ ⎛ ⎞
b. No, because + 1 has = ⎜ 2 lim x 2 + lim 1⎟ ⋅ ⎜ 7 lim x 2 + lim 13 ⎟ 8,1
x – 4x + x 2 + x + 6
4 3
⎝ x→ 2 x→ 2 ⎠ ⎝ x→ 2 x→ 2 ⎠
an asymptote at x ≈ 3.49.
⎡ ⎛ ⎞
2 ⎤⎡ ⎛ ⎞
2 ⎤
= ⎢2⎜ lim x ⎟ + 1⎥ ⎢7⎜ lim x ⎟ + 13⎥ 2
1 ⎢ ⎝ x→ 2 ⎠ ⎥⎢ ⎝ x→ 2 ⎠ ⎥
c. If δ ≤ , then 2.75 < x < 3 ⎣ ⎦⎣ ⎦
4
or 3 < x < 3.25 and by graphing = [2( 2 ) 2 + 1][7( 2 ) 2 + 13] = 135
x3 − x 2 − 2 x − 4
y = g ( x) =
x 4 − 4 x3 + x 2 + x + 6
on the interval [2.75, 3.25], we see that
x3 – x 2 – 2 x – 4
0< <3
x 4 – 4 x3 + x 2 + x + 6
so m must be at least three.
72 Section 1.3 Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.