Calculo
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
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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...
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