Question

$ \displaystyle \lim_{x\to0} \left\{\frac{1+tan\,x}{1+sin\,x}\right\}^{cosec\,x} $ is equal to

• A. $ \frac{1}{e} $
• B. $ 1 $
• C. $ e $
• D. $ e^{2} $

Answer 1

Answer: (B)

$\lim _{x \rightarrow 0}\left\{\frac{1+\tan x}{1+\sin x}\right\}^{\operatorname{cosec} x}$
$=\lim _{x \rightarrow 0} \frac{\left[\left.\left(1+\frac{\sin x}{\cos x}\right)^{\frac{\cos x}{\sin x}}\right|^{1 / \cos x}\right.}{(1+\sin x)^{1 / \sin x}}$
$=\frac{e^{\lim _{x \rightarrow 0} \frac{1}{\cos x}}}{e}=\frac{e}{e}=1$