Question

$\underset{x\to 0}{lim}{\left\{\frac{1+\tan x}{1+\sin x}\right)}^{\text{cosec}x}$ is equal to

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

Answer 1

The correct option is B $1$

$\underset{x\to 0}{lim}{\left\{\frac{1+\tan x}{1+\sin x}\right\}}^{\text{cosec}x}$
$=\underset{x\to 0}{lim}\left[\frac{{\left[{(1+\frac{\sin x}{\cos x})}^{\frac{\cos x}{\sin x}}\right]}^{\frac{1}{\cos x}}}{(1+\sin x{)}^{\frac{1}{\sin x}}}\right]$
$=\frac{e^{\underset{x\to 0}{lim}\frac{1}{\cos x}}}{e}=\frac{e}{e}=1$