Question
$\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} =$
दोनों पक्षों का $log$ लेने पर,
==> $\log y = \mathop {\lim }\limits_{x \to 0} \,\,\cot x\log \cos x$
==> $\log y = \mathop {\lim }\limits_{x \to 0} \frac{{\log \cos x}}{{\tan x}}$, $\left( {\frac{0}{0}} \right)$ रूप
$L-$ हॉस्पीटल नियम के प्रयोग से,
==> $\log y = \mathop {\lim }\limits_{x \to 0} \frac{{ - \tan x}}{{{{\sec }^2}x}}$= 0
==> $y = {e^0}$ ==> $y = 1$.
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