Question

$\underset{x\to 0}{\lim }{\left\{\frac{1+tanx}{1+sinx}\right\}}^{cosecx}$ is equal to

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

Answer 1

Answer: (B)

Consider $\underset{x\to 0}{\lim }{\left\{\frac{1+tanx}{1+sinx}\right\}}^{cosecx}$
$=\underset{x\to 0}{\lim }\frac{{\left[{\left(1+\frac{sinx}{cosx}\right)}^{\frac{cosx}{sinx}}\right]}^{1/cosx}}{{\left(1+sinx\right)}^{1/sinx}}$
[$\because tanx=\frac{sinx}{cosx}$ and $cosecx=\frac{1}{sinx}$]
We know, $\underset{n\to 0}{\lim }{\left(1+\frac{1}{n}\right)}^n=e$
$\therefore =\underset{x\to 0}{\lim }\frac{{\left[{\left(1+\frac{sinx}{cosx}\right)}^{\frac{cosx}{sinx}}\right]}^{1/cosx}}{{\left(1+sinx\right)}^{1/sinx}}$
$=\underset{x\to 0}{\lim }\frac{{\left[{\left(1+\frac{1}{\frac{cosx}{sinx}}\right)}^{\frac{cosx}{sinx}}\right]}^{\frac{1}{cosx}}}{\left[{\left(1+\frac{1}{cosecx}\right)}^{cosecx}\right]}$
$=\frac{e^{\underset{x\to 0}{\lim }\frac{1}{cosx}}}{e}$
$=\frac{e}{e}=1.$