$\lim_{x ightarrow 0}\left\{ \frac{1 + an x}{1 + \sin x} ight\}^{cosec\ x}$ is equal to

Question

$\lim_{x ightarrow 0}\left\{ \frac{1 + 	an x}{1 + \sin x} ight\}^{cosec\ x}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\lim_{x ightarrow 0}\left\{ \frac{1 + 	an x}{1 + \sin x} ight\}^{cosec\ x}$
$= \lim_{x ightarrow 0}\frac{\left\lbrack \left( 1 + \frac{\sin x}{\cos x} ight)^{\frac{\cos x}{\sin x}} ightbrack^{1/\cos x}}{\left( 1 + \sin x ight)^{1/\sin x}}$
$= \frac{e^{\lim_{x ightarrow 0}\ \frac{1}{\cos x}}}{e}$
$= \frac{e}{e} = 1$

Q. $\lim_{x \rightarrow 0}\left\{ \frac{1 + \tan x}{1 + \sin x} \right\}^{cosec\ x}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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