Solucionario Purcel
Páginas: 33 (8098 palabras)
Publicado: 28 de febrero de 2013
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....
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....
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