SECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE
|||| 7.7
S Click here for solutions.
Find the limit. Use l’Hospital’s Rule where appropriate. If
there is a more elementary method, use it. If l’Hospital’s Rule
doesn’t apply, explain why.
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23. lim
x sin 3x
x sin 3x
24. lim
e 4x 1
cos x
xl0
xl0
x2
1. lim 2
xl2 x 4
x 2 3x 4
2. lim
xl1
x1
25. lim
tan x sin x
x3
26. lim
x tan 2x
x tan 2x
x6 1
3. lim 4
x l 1 x 1
tan x
4. lim
x l 0 x sin x
27. lim
tan 2x
tanh 3x
28. lim
2x sin1x
2x cos1x
ex 1
5. lim
x l 0 sin x
x tan x
6. lim
xl0
sin x
29. lim
sin x
7. lim
xl0
x3
tan x
8. lim
xl
x
cos x
9. lim
x l32 x 32
11. lim
xla
3
3
xs
a
s
xa
4
t2
s
10. lim
t l16 t 16
12. lim
xl0
6x 2x
x
ln x3
13. lim
xl
x2
sin x
14. lim
xl0
ex
tan x
15. lim
xl0
x
2
sin x
16. lim
2
x l 0 tanx ln ln x
17. lim
xl
sx
ln1 e x
18. lim
xl
5x
tan12x
19. lim
xl0
3x
x
20. lim
1
x l 0 sin 3x
21. lim
xl0
1
Indeterminate Forms and L’Hospital’s Rule
A Click here for answers.
1–45
❙❙❙❙
sin mx
sin nx
xl0
xl0
xl0
xl0
1
xl0
2x sin x
2x tan1x
30. lim xe x
x l 31. lim e x ln x
32.
33. lim sx sec x
34. lim x cot x
35. lim x 1 tan x2
36. lim
xl
xl
xl0
xl0
xl1
xl0
sin10 x
sinx 10 sec 7x cos 3x
1
1
2
x4
x
37. lim ( x sx 2 1 )
xl
38. lim (sx 2 x 1 sx 2 x )
xl
39. lim
xl
40. lim
xl
1
1
x
x
e 1
x3
x3
x2 1
x2 1
41. lim sin xtan x
42. lim
43. lim cot xsin x
44. lim
xl
xl0
22. lim
lim
x l 2
xl
xl0
45. lim ln x
1
1
x2
1
1
x
x
x2
x
xl0
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2
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SECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE
Answers
E Click here for exercises.
1
4
3
3.
2
1.
S Click here for sol utions.
2. 5
4.
1
2
5.
1
6. 2
7.
∞
8. 0
9.
1
10.
11.
1
3a2/3
12. ln 3
13.
0
14. 0
15.
α
16. 1
17.
0
18.
2
3
m
21.
n
19.
23.
−2
1
2
2
27.
3
1
29.
3
25.
1
32
1
5
1
20.
3
22. 1
24. 0
26. −3
28. 0
30. 0
3
7
31.
0
32.
33.
0
34. 1
2
35. −
π
36. ∞
37.
0
38. 1
39.
0
40. 0
41.
1
42. 1
43.
1
44. ∞
45.
1
SECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE
❙❙❙❙
Solutions
E Click here for exercises.
A Click here for answers.
H
=.
1
x−2
x−2
1
= lim
= lim
=
1. lim 2
x→2 x − 4
x→2 (x − 2) (x + 2)
x→2 x + 2
4
The use of l’Hospital’s Rule is indicated by an H above the equal sign:
2.
x2 + 3x − 4
(x − 1) (x + 4)
= lim
x→1
x→1
x−1
x−1
= lim (x + 4) = 5
lim
16.
17.
x→1
6
3.
−6
3
x −1 H
6x5
=
lim
=
=
x→−1 x4 − 1
x→−1 4x3
−4
2
4.
1
1
tan x H
sec2 x
=
=
= lim
x→0 x + sin x
x→0 1 + cos x
1+1
2
5.
1
ex − 1 H
ex
= lim
= =1
x→0 sin x
x→0 cos x
1
lim
lim
lim
1 + 12
x + tan x H
1 + sec2 x
=
=2
= lim
x→0
x→0
sin x
cos x
1
sin x H
cos x
7. lim
= lim
=∞
x→0 x3
x→0 3x2
tan π
0
tan x
=
= = 0. L’Hospital’s Rule does not
8. lim
x→π
x
π
π
apply because the denominator doesn’t approach 0.
6.
lim
cos x H
− sin x
= − sin 3π
= lim
=1
2
x − 3π/2 x→3π/2
1
√
√
4
4
t−2
t−2
√
10. lim
= lim √
t→16 t − 16
t→16
t+4
t−4
√
4
t−2
= lim √
√
√
4
t→16
t+4
t+2 4t−2
9.
11.
12.
15.
19.
20.
1
1√
1 − 9x2 =
3
3
m
sin mx H
m cos mx
=
21. lim
= lim
x→0 sin nx
x→0 n cos nx
n
= lim
x→0
22.
lim
lim
x→0
x→3π/2
lim
x→a
23.
24.
1
1
=
=
(4 + 4) (2 + 2)
32
25.
lim
1+3
x + sin 3x H
1 + 3 cos 3x
=
= −2
= lim
x − sin 3x x→0 1 − 3 cos 3x
1−3
lim
0
e4x − 1
= =0
cos x
1
x→0
x→0
lim
x→0
1
x1/3 − a1/3 H
(1/3) x−2/3
= lim
= 2/3
x→a
x−a
1
3a
6x − 2x H
6x (ln 6) − 2x (ln 2)
= lim
x→0
x→0
x
1
= ln 6 − ln 2 = ln 62 = ln 3
lim
lim
tan αx H
α sec2 αx
=α
= lim
x→0
x→0
x
1
lim
sin10 x H
10 sin9 x cos x
=
lim
x→0 10x9 cos (x10 )
sin (x10 )
9
sin x
cos x
= lim
lim
x→0
x→0 cos (x10 )
x
= 19 · 1 = 1
1
√
= lim √
t→16
t+4 4t+2
(ln x)3 H
3 (ln x)2 (1/x)
3 (ln x)2
= lim
= lim
2
x→∞
x→∞
x→∞
x
2x
2x2
6 (ln x) (1/x)
3 ln x
H
= lim
= lim
x→∞
x→∞ 2x2
4x
3/x
3
H
= lim
= lim
=0
x→∞ 4x
x→∞ 4x2
0
sin x
14. lim
= = 0. L’Hospital’s Rule does not apply.
x→0 ex
1
13.
18.
sin2 x H
2 sin x cos x
= lim
x→0 tan (x2 )
x→0 2x sec2 (x2 )
sin x
cos x
= lim
lim
=1·1= 1
x→0
x x→0 sec2 (x2 )
1 (x ln x)
ln ln x H
2
lim √
= lim √
=0
= lim √
x→∞
x→∞ 1 (2 x)
x→∞
x
x ln x
ex (1 + ex )
ln (1 + ex ) H
lim
= lim
x→∞
x→∞
5x
5
1
ex
ex
H
=
lim
=
= lim
x→∞ 5 (1 + ex )
x→∞ 5ex
5
2
2 1 + 4x
2
tan−1 (2x) H
lim
= lim
=
x→0
x→0
3x
3
3
x
1
H
lim
= lim x→0 sin−1 (3x)
x→0
3
1 − (3x)2
lim
26.
lim
x→0
tan x − sin x H
sec2 x − cos x
= lim
x→0
x3
3x2
2
2 sec x tan x + sin x
H
= lim
x→0
6x
4 sec2 x tan2 x + 2 sec4 x + cos x
H
= lim
x→0
6
0+2+1
1
=
=
6
2
x + tan 2x H
1 + 2 sec2 2x
= lim
x − tan 2x x→0 1 − 2 sec2 2x
=
27.
28.
lim
x→0
1 + 2 (1)2
= −3
1 − 2 (1)2
tan 2x H
2 sec2 2x
2
= lim
=
tanh 3x x→0 3 sech2 3x
3
2 (0) − 0
2x − sin−1 x
=
= 0. L’Hospital’s Rule
x→0 2x + cos−1 x
2 (0) + π/2
does not apply.
lim
3
4
❙❙❙❙
SECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE
29.
30.
√
2x − sin−1 x H
2 − 1/ 1 − x2
=
lim
x→0 2x + tan−1 x
x→0 2 + 1/ (1 + x2 )
2−1
1
=
=
2+1
3
lim
= lim
x→−∞
=0
lim e−x ln x = lim
lim
x→(π/2)−
x→0+
lim (x − π) cot x = lim
x→π
x→π
=
x−π H
1
= lim
x→π sec2 x
tan x
1
=1
(−1)2
x−1
cot (πx/2)
2
1
H
= lim
π =−
π
x→1+ − csc2 (πx/2) 2
lim (x − 1) tan (πx/2) = lim
x→1+
x→1+
1
1
1 − x2
− 2 = lim
=∞
4
x→0
x→0
x
x
x4
√
37. lim x − x2 − 1
36.
41.
y = (sin x)tan x
⇒ ln y = tan x ln (sin x), so
ln (sin x)
lim ln y = lim tan x ln (sin x) = lim
cot x
x→0 +
x→0+
x→0 +
(cos
x)
/
sin
x
H
= lim
− csc2 x
x→0+
= lim (− sin x cos x) = 0 ⇒
x→0+
tan x
= lim eln y = e0 = 1.
lim (sin x)
x→0 +
x→0 +
x
1
1
⇒
42. Let y = 1 + 2
. Then ln y = x ln 1 + 2
x
x
1
ln 1 + 2
x
1
lim ln y = lim x ln 1 + 2 = lim
x→∞
x→∞
x→∞
x
1/x
2
1
− 3
1+ 2
x
x
H
= lim
x→∞
−1/x2
2/x
= 0,
1 + 1/x2
x
so lim 1 + 1/x2 = lim eln y = e0 = 1.
= lim
x→∞
lim
x→∞
43.
x→∞
√
√
x + x2 − 1
2
√
= lim x − x − 1
x→∞
x + x2 − 1
2
2
x − x −1
√
= lim
x→∞ x +
x2 − 1
1
√
=0
= lim
x→∞ x +
x2 − 1
√
√
38. lim
x2 + x + 1 − x2 − x
x→∞
√
√
√
x2 + x + 1 + x2 − x
√ 2
2
√
x +x+1− x −x √
= lim
x→∞
x2 + x + 1 + x2 − x
2
2
x +x+1 − x −x
= lim √
√
x→∞
x2 + x + 1 + x2 − x
2x + 1
√
= lim √ 2
x→∞
x + x + 1 + x2 − x
2 + 1/x
= lim x→∞
1 + 1/x + 1/x2 + 1 − 1/x
2
=1
=
1+1
1
1
1
1
39. lim
− x
= lim
− lim x
(since
x→∞
x→∞ x
x→∞ e − 1
x e −1
both limits exist) = 0 − 0 = 0
2x3
2/x
= lim
− 1 x→∞ 1 − 1/x4
x→∞ x4
3 (−1)
3
−3 sin 3x
=
=
−7 sin 7x
7 (−1)
7
√
33. lim
x sec x = 0 · 1 = 0
x3 x2 + 1 − x3 x2 − 1
= lim
x→∞
(x2 − 1) (x2 + 1)
x H
1
= lim
x→−∞ −e−x
e−x
= lim −ex = 0
H
35.
x→∞
x3
x3
−
x2 − 1 x2 + 1
x→−∞
=
34.
lim
lim xex = lim
x→−∞
ln x H
1/x
= lim x
x→∞
x→∞ ex
x→∞ e
1
= lim
=0
x→∞ xex
cos 3x
32.
lim sec 7x cos 3x = lim
x→(π/2)−
x→(π/2)− cos 7x
31.
40.
x→∞
sin x
y = (cot x)
⇒
ln y = sin x ln (cot x) ⇒
− csc2 x cot x
ln (cot x) H
= lim
lim ln y = lim
csc x
− csc x cot x
x→0 +
x→0+
x→0 +
csc x
sin x
= lim
= lim
=0
2
x→0+ cot x
x→0 + cos2 x
so lim (cot x)sin x = lim eln y = e0 = 1.
x→0+
44.
x→0+
x2
Let y = (1 + 1/x) . Then ln y = x2 ln (1 + 1/x) ⇒
ln (1 + 1/x)
lim ln y = lim x2 ln (1 + 1/x) = lim
x→∞
x→∞
x→∞
1/x2
−1/x2 / (1 + 1/x)
H
= lim
x→∞
−2/x3
x
=∞ ⇒
= lim
x→∞ 2 (1 + 1/x)
2
lim (1 + 1/x)x = lim eln y = ∞.
x→∞
45.
x→∞
x
y = (− ln x)
⇒ ln y = x ln (− ln x), so
ln (− ln x)
lim ln y = lim x ln (− ln x) = lim
1/x
x→0 +
x→0+
x→0+
(1/ − ln x) (−1/x)
−1/x2
−x
=0 ⇒
= lim
x→0+ ln x
lim (− ln x)x = e0 = 1.
H
= lim
x→0+
x→0 +