Question

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

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

Answer 1

Answer: (C)

$1$

$\underset{x\to 0}{lim}{\left(\frac{1+\tan x}{1+\sin x}\right)}^{\text{cosec}x}=\underset{x\to 0}{lim}{e}^{(\frac{1+\tan x}{1+sinx}-1)\text{cosec}x}=\underset{x\to 0}{lim}{e}^{\left(\frac{\tan x-\sin x}{1+\sin x}\right)\text{cosec}x}=\underset{x\to 0}{lim}{e}^{\left(\frac{\sec x-1}{1+\sin x}\right)}=e^0=1$