Question

$\displaystyle\lim_{x \to 0}\left\{\frac{1+tan \,x}{1+sin \,x}\right\}^{cosec \,x} $ is equal to

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

Answer 1

Answer: (B)

Consider $\displaystyle\lim_{x \to 0}\left\{\frac{1+tan \,x}{1+sin \,x}\right\}^{cosec \,x} $
$= \displaystyle\lim _{x \to 0} \frac{\left[\left(1+\frac{sin \,x}{cos\, x}\right)^{\frac{cos \,x}{sin\, x}}\right]^{1/cos \,x}}{\left(1+sin x\right)^{1/ sin \,x}}$
[$\because tan\, x = \frac{sin \,x}{cos \,x}$ and $cosec\, x = \frac{1}{sin \,x}$]
We know, $\displaystyle\lim _{n \to 0}\left(1+\frac{1}{n}\right)^{n} = e$
$\therefore = \displaystyle\lim _{x \to 0} \frac{\left[\left(1+\frac{sin\, x}{cos\, x}\right)^{\frac{cos\, x}{sin \,x}}\right]^{1/cos \,x}}{\left(1+sin\, x\right)^{1/ sin \,x}}$
$= \displaystyle\lim _{x \to 0} \frac{\left[\left(1+\frac{1}{\frac{cos \,x}{sin \,x}}\right)^{\frac{cos \,x}{sin\, x}}\right]^{\frac{1}{cos \,x}}}{\left[\left(1+\frac{1}{cosec \,x}\right)^{cosec\, x}\right]}$
$= \frac{e^{\displaystyle\lim_{x \to 0} \frac{1}{cos \,x}}}{e}$
$ = \frac{e}{e} = 1.$