Question

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

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

Answer 1

Answer: (C)

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