# Solucionario 1 capitulo de purcell

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CHAPTER

1
9.
lim = lim
x → –1 2

Limits
x3 – 4 x 2 + x + 6 x → –1 x +1 ( x + 1)( x 2 – 5 x + 6) x → –1 x +1

1.1 Concepts Review
1. L; c 2. 6 3. L; right 4. lim f ( x) = M
x →c

= lim ( x 2 – 5 x + 6) = (–1) – 5(–1) + 6 = 12

Problem Set 1.1
1. lim( x – 5) = –2
x →3

10. lim

x 4 + 2 x3 – x 2

x2 = lim( x 2 + 2 x –1) = –1
x →0 x →0

2. lim (1 – 2t ) = 3
t → –1

11.3. 4.

x →−2

lim ( x 2 + 2 x − 1) = (−2) 2 + 2(−2) − 1 = −1
lim ( x 2 + 2t − 1) = (−2) 2 + 2t − 1 = 3 + 2t

x2 – t 2 ( x + t )( x – t ) = lim x→–t x + t x→ – t x+t = lim ( x – t ) lim
x→ –t

= –t – t = –2t
12.
x2 – 9 x →3 x – 3 ( x – 3)( x + 3) = lim x →3 x–3 = lim( x + 3) lim
x →3

x →−2

5. lim t 2 − 1 =
t →−1

( (

) ( ( −1) − 1) = 0
2

6. lim t 2 − x 2 =
t →−1) ( ( −1)

2

− x2 = 1 − x2

)

=3+3=6
13.
lim (t + 4)(t − 2) 4 (3t − 6) 2 (t − 2) 2 t + 4 9(t − 2) 2
t+4 9

7. lim

x2 – 4 ( x – 2)( x + 2) = lim x→2 x – 2 x→2 x–2 = lim( x + 2)
x→2

t →2

=2+2=4
8.
t 2 + 4t – 21 t → –7 t+7 (t + 7)(t – 3) = lim t → –7 t+7 = lim (t – 3) lim
t → –7

= lim
= lim =

t →2

t →2

2+4 6 = 9 9

= –7 – 3 = –10

14.

t →7+

lim

(t− 7)3 t −7
+

= lim
t →7

(t − 7) t − 7 t −7 t −7

= lim

t →7+

= 7−7 = 0

Instructor’s Resource Manual

Section 1.1

63

© 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 inwriting from the publisher.

15. lim

x 4 –18 x 2 + 81 ( x – 3)
2

x →3

= lim

( x 2 – 9) 2 ( x – 3)
2

x →3

lim

t →0

1 − cos t =0 2t

= lim = 36

( x – 3) 2 ( x + 3) 2 ( x – 3)
2

x →3

= lim( x + 3)2 = (3 + 3) 2
x →3

21.

x 1. 0.1 0.01 0.001 –1. –0.1 –0.01 –0.001
x →0

( x − sin x ) 2 / x 2

0.0251314
2.775 × 10−6 2.77775 × 10−10 2.77778 × 10−14

16.lim

(3u + 4)(2u – 2)3 (u –1) 2

u →1
u →1

= lim

8(3u + 4)(u –1)3 (u –1) 2

u →1

= lim 8(3u + 4)(u – 1) = 8[3(1) + 4](1 – 1) = 0 (2 + h) 2 − 4 4 + 4h + h 2 − 4 = lim h→0 h→0 h h lim = lim h 2 + 4h = lim(h + 4) = 4 h →0 h →0 h

0.0251314 2.775 × 10−6
2.77775 × 10−10 2.77778 × 10−14 =0
(1 − cos x ) / x
2 2

17.

lim

( x – sin x) 2 x2

18.

( x + h) 2 − x 2 x 2 + 2 xh +h 2 − x 2 = lim h→0 h →0 h h lim = lim h 2 + 2 xh = lim(h + 2 x) = 2 x h →0 h →0 h

22.

x 1. 0.1 0.01 0.001 –1. –0.1 –0.01 –0.001
x →0

0.211322 0.00249584 0.0000249996 2.5 × 10−7 0.211322 0.00249584 0.0000249996 2.5 × 10−7
=0

19.

x 1. 0.1 0.01 0.001 –1. –0.1 –0.01 –0.001
lim

sin x 2x

0.420735 0.499167 0.499992 0.49999992 0.420735 0.499167 0.499992 0.49999992
23.

lim(1 – cos x) 2 x2

t 2. 1.1 1.01 1.001 0 0.9 0.99 0.999
2

(t − 1) /(sin(t − 1))

2

3.56519 2.1035 2.01003 2.001 1.1884 1.90317 1.99003 1.999

sin x = 0.5 x →0 2 x
20.

t 1. 0.1 0.01 0.001 –1. –0.1 –0.01 –0.001

1− cos t 2t

0.229849 0.0249792 0.00249998 0.00024999998 –0.229849 –0.0249792 –0.00249998 –0.00024999998

lim

t →1 sin(t

t −1 =2 − 1)

64

Section 1.1Instructor’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.

27. 24.

x
1. + π 4 0.1 +

( x − π / 4) 2 /(tan x − 1) 2

x 4. 3.1 3.01 3.001 2.2.9 2.99 2.999

x −sin( x − 3) − 3 x −3

0.0320244
π
4

0.158529 0.00166583 0.0000166666 1.66667 × 10−7 0.158529 0.00166583 0.0000166666 1.66667 × 10−7

0.201002
π
4

0.01 +

0.245009
π
4

0.001 + −1. + π 4

0.2495 0.674117 0.300668 0.255008 0.2505
= 0.25
(2 − 2sin u ) / 3u

−0.1 + π 4

−0.01 + π 4

−0.001 + π 4
x→ π
4

x – sin( x – 3) – 3 =0 lim x →3 x–3
25....