Question

The value of $\underset{x\to 0^{+}}{\mathrm{Lim}}\left(x^x+(\tan x{)}^{\mathrm{cosec}x}+(\mathrm{cosec}x{)}^{\tan x}\right)$ is equal to

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

Answer 1

Answer: (C)

Let $\underset{x\to 0^{+}}{\mathrm{Lim}}\left({\left(x^x+(\tan x{)}^{\mathrm{cosec}x}+(\mathrm{cosec}x)\right)}^{\tan x}\right)$
$=l_1+l_2+l_3$, where $l_1=\underset{x\to 0^{+}}{\mathrm{Lim}}{x}^x\left(0^{\circ }\right)$
$\Rightarrow \ln l_1=\underset{x\to 0^{+}}{\mathrm{Lim}}x\cdot \ln x(0\times \infty )=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{\ln x}{\frac{1}{x}}\left(\frac{\infty }{\infty }\right)=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{\frac{1}{x}}{\frac{-1}{x^2}}=0\Rightarrow l_1=1$.
$\text{ Now, }{l}_2=\underset{x\to 0^{+}}{\mathrm{Lim}}(\tan x{)}^{\mathrm{cosec}x}=0\left(0^{\infty }\right)$
$\text{ Also, }{l}_3=\underset{x\to 0^{+}}{\mathrm{Lim}}(\mathrm{cosec}x{)}^{\tan x}\left({\infty }^0\right)$
$\Rightarrow \ln l_3=\underset{x\to 0^{+}}{\mathrm{Lim}}(\tan x)\cdot \ln (\mathrm{cosec}x)(0\times \infty )$
$=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{\ln (\mathrm{cosec}x)}{\cot x}\left(\frac{\infty }{\infty }\right)$
$=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{\frac{1}{\mathrm{cosec}x}(-\mathrm{cosec}x\cdot \cot x)}{-{\mathrm{cosec}}^{2}x}=\underset{x\to 0^{+}}{\mathrm{Lim}}\left(\frac{{\sin }^{2}x}{\tan x}\right)=0\to l_3=1$
Hence, $\left(l_1+l_2+l_3\right)=1+0+1=2$.