Question

Value of $\underset{x\to 0}{lt}{\left(\frac{1+\tan x}{1+\sin x}\right)}^{cosecx}$ equals.

• A. $1$
• B. $e$
• C. $\frac{1}{e}$
• D. None of these

Answer 1

Answer: (A)

$1$

$\underset{x\to 0}{lt}{\left(\frac{1+\tan x}{1+\sin x}\right)}^{cosecx}\left(1^{\infty }form\right)$
$=e^{\underset{x\to 0}{lt}\left(\frac{1+\tan x-(1+\sin x)}{1+\sin x}\right)\frac{1}{\sin x}}$
(Using $\underset{x\to 0}{lt}f(x{)}^{g\left(x\right)}=e^{\underset{x\to 0}{lt}f\left(x\right)-1\left)g\right(x)}$)
$=e^{\underset{x\to 0}{lt}\left(\frac{\tan x-\sin x}{1+\sin x}\right)\left(\frac{1}{\sin x}\right)}$
$=e^{\underset{x\to 0}{lt}\frac{1-\cos x}{(1+\sin x)\cos x}}$

$=e^0$

$=1$.
Hence choice (a) is correct answer.