1. Solutions to Worksheet for Section 3.7
Indeterminate Forms and L’Hˆpital’s Rule
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V63.0121, Calculus I
Summer 2010
Evaluate the following limits.
x cos x
1. lim
x→0 sin x + 2x
Solution. We have
x cos x H cos x − x sin x 1
lim = lim =
x→0 sin x + 2x x→0 cos x + 2 3
sin2 x
2. lim
x→0 x sin x + 3x2
Solution. We have to use l’Hˆpital’s Rule twice:
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sin2 x H 2 sin x cos x
lim = lim
x→0 x sin x + 3x2 x→0 x cos x + sin x + 6x
H 2 cos2 x − 2 sin2 x
= lim
x→0 cos x − x sin x + cos x + 6
2 1
= = .
8 4
sin 4x
3. lim
x→0 cos 2x + 1
Solution. This is a red herring. The numerator goes to 0 while the denominator goes to 2.
So the limit is 0.
4. lim 4x2 + x − 2x
x→∞
Solution. Multiply by the conjugate radical:
√
4x2 + x + 2x x
lim 4x2 + x − 2x = lim 4x2 + x − 2x √ = lim √
x→∞ x→∞ 4x 2 + 1 + 2x x→∞ 4x 2 + 1 + 2x
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2. At this point l’Hˆpital’s Rule isn’t needed either:
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x x 1
lim √ = lim =
x→∞ 4x2 + 1 + 2x x→∞ x 4 + 1/x2 + 2 4
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5. lim x3 e−x
x→∞
Solution. We can use multiple applications of l’Hˆpital’s Rule, but with some judicious can-
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celation as well.
2 x3 H 3x2
lim x3 e−x = lim x2
= lim 2
x→∞ x→∞ e x→∞ 2xex
3x H 3
= lim 2 = lim 2 = 0
x→∞ 2ex x→∞ 4xex
6. lim+ xsin 2x
x→0
Solution. This limit is of the form 00 . Let the limit above be L. Then
ln L = ln lim xsin 2x = lim ln xsin 2x = lim sin 2x ln x
x→0+ x→0+ x→0+
ln x H 1/x tan 2x
= lim+ = lim+ = −2 lim+ sin 2x ·
x→0 csc 2x x→0 −2 csc 2x cot 2x x→0 x
H 2 sec2 2x
= −2 · 0 · lim+ = −2 · 0 · 2 = 0
x→0 1
So L = e0 = 1.
7. lim (ln x)1/x
x→∞
Solution. This limit is of the form ∞0 . Let it be L. Then
ln x H 1/x
ln L = lim = lim =0
x→∞ x x→∞ 1
So L = e0 = 1.
8. lim (cos x)tan x
x→π/2−
Solution. This limit is of the form 1∞ . Let it be L. Then
ln(sin x)
ln L = lim tan x ln(sin x) = lim
x→π/2 − x→π/2 − cot x
− cos x
H sin x sin2 x cos x
= lim − = lim − = 1 · 0 = 0.
x→π/2 − csc2 x x→π/2 sin x
0
So L = e = 1.
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