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$\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$ is equal to :
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Verified Answer
The correct answer is:
$\frac{1}{e}$
Let $y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$
Taking $\log$ on both sides, we get
$\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x}\left[\frac{\infty}{\infty}\right.$ form $]$
$=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad$ (ByL' Hospital rule)
$=-\lim _{x \rightarrow 0} \frac{x}{\tan x} \quad\left(\because \cot x=\frac{1}{\tan x}\right)$
$\Rightarrow \log y=-1$
$\Rightarrow y=e^{-1}=\frac{1}{e}$
Hence, required limit $=\frac{1}{e}$
Taking $\log$ on both sides, we get
$\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x}\left[\frac{\infty}{\infty}\right.$ form $]$
$=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad$ (ByL' Hospital rule)
$=-\lim _{x \rightarrow 0} \frac{x}{\tan x} \quad\left(\because \cot x=\frac{1}{\tan x}\right)$
$\Rightarrow \log y=-1$
$\Rightarrow y=e^{-1}=\frac{1}{e}$
Hence, required limit $=\frac{1}{e}$
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