Question

The value of $\underset{x\to 0}{lim}{\left\{\frac{1+\tan x}{1+\sin x}\right\}}^{\csc x}$ equals

• A. $\theta$
• B. $\frac{1}{e}$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

${lim}_{x\to 0}{\left\{\frac{1+\tan x}{1+\sin x}\right\}}^{cosecx}=?$

Let $y={\left\{\frac{1+\tan x}{1+\sin x}\right\}}^{cosecx}$

$\log y=cosecx\log \left\{\frac{1+\tan x}{1+\sin x}\right\}$

$\log y=cosecx[\frac{1+\tan x}{1+\sin x}-1]$

$\underset{x\to 0}{lim}y=e\underset{x\to 0}{lim}cosecx\left[\frac{1+\tan x-1-\sin x}{1+\sin x}\right]$

$\underset{x\to 0}{lim}y=e\underset{x\to 0}{lim}cosecx\left[\frac{\tan x-\sin x}{1+\sin x}\right]$

$\underset{x\to 0}{lim}y=e\underset{x\to 0}{lim}cosecx\sin x\left[\frac{\sec x-1}{1+\sin x}\right]$

$\underset{x\to 0}{lim}y=e\underset{x\to 0}{lim}\left[\frac{\sec x-1}{1+\sin x}\right]$ where $cosecx-\sin x=1$

$\underset{x\to 0}{lim}y=e^0=1$

$\underset{x\to 0}{lim}{\left\{\frac{1+\tan x}{1+\sin x}\right\}}^{cosecx}=1$