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Q. The value of $\displaystyle\lim _{x \rightarrow \pi / 2}(\sec x)^{\cot x}$ is

Limits and Derivatives

Solution:

$y=\displaystyle\lim _{x \rightarrow \pi / 2}(\sec x)^{\cot x} \,\,\,\left(\infty^{0}\right.$ form)
$\therefore \log y=\displaystyle\lim _{x \rightarrow \pi / 2} \cot x \log \sec x$
$=\displaystyle\lim _{x \rightarrow \pi / 2} \frac{\log \sec x}{\tan x} \,\,\,(\frac{\infty}{\infty}$ form $)$
$=\displaystyle\lim _{x \rightarrow \pi / 2} \frac{1}{\sec x} \times \frac{\sec x \tan x}{\sec ^{2} x}$ (Using L'Hospital's Rule)
$=\displaystyle\lim _{x \rightarrow \pi / 2} \frac{\tan x}{\sec ^{2} x}=\displaystyle\lim _{x \rightarrow \pi / 2} \sin x \cos x=0$
$\therefore y=e^{0}=1$