# Calculo

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CHAPTER

8

Indeterminate Forms and Improper Integrals
7. The limit is not of the form
0 . 0

8.1 Concepts Review
1. lim f ( x); lim g ( x)
x→a x →a

As x → 1– , x 2 – 2 x + 2 → 1, and x 2 – 1 → 0 – so
x →1–

2.

f ′( x) g ′( x)
x →0

lim

x2 – 2 x + 2 x2 + 1

= –∞

3. sec2 x; 1; lim cos x ≠ 0 4. Cauchy’s Mean Value

8. The limit is of the form
lim ln x 2

0 . 0Problem Set 8.1
0 1. The limit is of the form . 0 2 x – sin x 2 – cos x = lim =1 lim x 1 x →0 x →0
2. The limit is of the form
0 . 0 cos x – sin x lim = lim =1 x →π / 2 π / 2 – x x →π / 2 –1 0 . 0 1 – 2 cos 2 x
sec2 x

1 2x 2 1 = lim x = lim =1 x →1 x 2 – 1 x →1 2 x x →1 x 2

9. The limit is of the form

0 . 0
1

3 ln(sin x)3 = lim sin x lim x →π / 2 π / 2 – x x →π / 2 0 = =0 –13sin 2 x cos x –1

10. The limit is of the form
lim

0 . 0

3. The limit is of the form
x – sin 2 x = lim x →0 tan x x →0 lim

ex – e– x ex + e– x 2 = lim = =1 2 x →0 2sin x x →0 2 cos x

=

1– 2 = –1 1

11. The limit is of the form

0 . 0

4. The limit is of the form
lim tan –1 3 x sin –1 x =

0 . 0
= 3 =3 1

1 – 2t –3 t – t2 3 2 t = lim = 2 =– lim 1 1 2 t →1 ln t t →1t

x →0

3 1+ 9 x 2 lim x →0 1 1– x 2

12. The limit is of the form
7 2
x x

0 . 0
= lim
x →0

0 5. The limit is of the form . 0

x →0

lim

–1 –1

+

= lim

7 x ln 7 2 x 2 x

7 2

x x

ln 7 ln 2

x →0+ 2 x ln 2

+

– 3x –10 2 2 = =– –7 7

x → –2 x 2

lim

x2 + 6 x + 8

= lim

2x + 6 x → –2 2 x – 3

=

ln 7 ≈ 2.81 ln 2

13. The limit is of theform

0 . (Apply l’Hôpital’s 0
= lim = –2sin 2 x

0 6. The limit is of the form . 0
x →0

Rule twice.)
x →0

lim

ln cos 2 x 7x
2

= lim

–2sin 2 x cos 2 x

lim

x3 – 3 x 2 + x x3 – 2 x

= lim

3x2 + 6 x + 1 3x2 – 2

x →0

=

1 1 =– –2 2

x →0

14 x

x →0 14 x cos 2 x

= lim

–4 cos 2 x

x →0 14 cos 2 x – 28 x sin 2 x

–4 2 =– 14 – 0 7

476Section 8.1

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.

14. The limit is of the form

0 . 0 3sin x 3cos x lim = lim 1 –x x →0 – x→0 – –
2 –x x →0 –

19. The limit is of the form

0 . (Apply l’Hôpital’s 0 –1
–2 x (1+ x 2 ) 2

Rule twice.)
24 x 2 1 1 = lim – =– 24 x →0 24(1 + x 2 ) 2
x →0 x →0

= lim – 6 – x cos x = 0

lim

tan –1 x – x 8 x3

= lim

1 1+ x 2

= lim

x →0

48 x

15. The limit is of the form

0 . (Apply l’Hôpital’s 0

Rule three times.) tan x – x sec2 x – 1 lim = lim x →0 sin 2x – 2 x x →0 2 cos 2 x – 2
2sec 2 x tan x 2sec 4 x + 4sec2 x tan 2 x = lim –8cos 2 x x →0 –4sin 2 x x →0 2+0 1 = =– –8 4 = lim

20. The limit is of the form

0 . (Apply l’Hôpital’s 0

Rule twice.) cosh x –1 sinh x cosh x 1 = lim = lim = lim 2` 2 2 x →0 x →0 2 x x →0 x
21. The limit is of the form
0 . (Apply l’Hôpital’s 0

0 16. The limit is of the form . (Apply l’Hôpital’s 0 Rule threetimes.) sin x – tan x cos x – sec2 x lim = lim x →0 x 2 sin x x →0 2 x sin x + x 2 cos x – sin x – 2sec2 x tan x = lim x →0 2sin x + 4 x cos x – x 2 sin x – cos x – 2sec4 x – 4sec2 x tan 2 x = lim x →0 6 cos x – x 2 cos x – 6 x sin x –1 – 2 – 0 1 = =– 6–0–0 2
17. The limit is of the form

Rule twice.) 1 − cos x − x sin x lim 2 + x → 0 2 − 2 cos x − sin x − x cos x = lim x → 0+ 2sin x − 2 cos xsin s x sin x – cos x = lim 2 2 + x →0 2 cos x – 2 cos x + 2sin x 0 This limit is not of the form . 0 As x → 0+ , x sin x – cos x → −1 and
2 cos x – 2 cos 2 x + 2sin 2 x → 0+ , so x sin x – cos x lim = –∞ + 2 cos x – 2 cos 2 x + 2sin 2 x x →0

0 . (Apply l’Hôpital’s 0

Rule twice.) x2 2x 2 lim = lim = lim + sin x – x + cos x – 1 + − sin x x →0 x →0 x →0 0 This limit is not of the form . As...