464
Chapter 4 | Applications of Derivatives
x sin x .
Evaluate lim x →0 +
Solution Let
y = x sin x .
Therefore,
ln y = ln ⎛
⎞ ⎠ = sin x ln x .
⎝ x sin x
We now evaluate lim x →0 +
sin x ln x . Since lim x →0 +
sin x =0 and lim x →0 +
ln x =−∞, we have the indeterminate
form 0·∞. To apply L’Hôpital’s rule, we need to rewrite sin x ln x as a fraction. We could write sin x ln x = sin x 1/ln x or sin x ln x = ln x 1/sin x = ln x csc x . Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain lim x →0 + sin x ln x = lim x →0 + sin x 1/ln x = lim x →0 + cos x −1/ ⎛ ⎝ x (ln x ) 2 ⎞ ⎠ = lim x →0 + ⎛ ⎝ − x (ln x ) 2 cos x ⎞ ⎠ . Unfortunately, we not only have another expression involving the indeterminate form 0·∞, but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing sin x ln x = ln x 1/sin x = ln x csc x , and applying L’Hôpital’s rule, we obtain lim x →0 + sin x ln x = lim x →0 + ln x csc x = lim x →0 + 1/ x −csc x cot x = lim x →0 + −1 x csc x cot x . Using the fact that csc x = 1 sin x and cot x = cos x sin x , we can rewrite the expression on the right-hand side as lim x →0 + −sin 2 x x cos x = lim x →0 + ⎡ ⎣ sin x x · (−tan x ) ⎤ ⎦ = ⎛ ⎝ lim x →0 + sin x x ⎞ ⎠ · ⎛ ⎝ lim x →0 + (−tan x ) ⎞ ⎠ =1·0=0.
ln y =0. Therefore, ln ⎛
y ⎞ ⎠ =0 and we have
⎝ lim
We conclude that lim x →0 +
x →0 +
x sin x = e 0 =1.
lim x →0 +
y = lim
x →0 +
Hence,
x sin x =1.
lim x →0 +
x x .
4.43
Evaluate lim x →0 +
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