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$\lim _{x \rightarrow 0} \frac{\log \cos x}{x}=$
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2981 Upvotes
Verified Answer
The correct answer is:
$0$
$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{\log (1-2 \sin^2 (\frac{x}{2})) }{x} \\
& =\lim _{x \rightarrow 0} \frac{-\left[2 \sin ^2 \frac{x}{2}+\left(\frac{2 \sin ^2 \frac{x}{2}}{2}\right)^2+\ldots \ldots .\right]}{x}=0
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{-\tan x}{1}=0 .$
& \lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{\log (1-2 \sin^2 (\frac{x}{2})) }{x} \\
& =\lim _{x \rightarrow 0} \frac{-\left[2 \sin ^2 \frac{x}{2}+\left(\frac{2 \sin ^2 \frac{x}{2}}{2}\right)^2+\ldots \ldots .\right]}{x}=0
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{-\tan x}{1}=0 .$
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