Question

The value of $\underset{x\to 0}{lim}{\left(\frac{1+\tan x}{1+\sin x}\right)}^{1/\sin x}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $\infty$

Answer 1

The correct option is A $1$

Let, $L=\underset{x\to 0}{lim}{\left(\frac{1+tanx}{1+sinx}\right)}^{1/sinx}$

Clearly form of the limit is $1^{\infty }$, So

$L=\underset{x\to 0}{lim}{(1+\frac{1+tanx}{1+sinx}-1)}^{1/sinx}$

$=\underset{x\to 0}{lim}{(1+\frac{tanx-sinx}{1+\sin x})}^{\frac{1+\sin x}{tanx-sinx}.\frac{tanx-sinx}{(1+\sin x)\sin x}}$

$=e^{\underset{x\to 0}{lim}\frac{tanx-sinx}{(1+\sin x)\sin x}}$

$=e^{\underset{x\to 0}{lim}\frac{\frac{tanx}{x}-\frac{sinx}{x}}{(1+\sin x)\frac{\sin x}{x}}}=e^{1-1}=e^0=1$