Question

The value of $\underset{x\to 0^{+}}{lim}[x\sin \left(\frac{1}{x}\right)+(\sin x{)}^{1/x}+{\left(\frac{1}{x}\right)}^{\sin x}],$ is

• A. $1$
• B. $0$
• C. $-1$
• D. Limit does not exist

Answer 1

The correct option is A $1$

$L=\underset{x\to 0^{+}}{lim}[x\sin \left(\frac{1}{x}\right)+(\sin x{)}^{1/x}+{\left(\frac{1}{x}\right)}^{\sin x}]$
$\Rightarrow L=0+0+\underset{x\to 0^{+}}{lim}{\left(\frac{1}{x}\right)}^{\sin x}$
$\Rightarrow L=\underset{x\to 0^{+}}{lim}{\left(\frac{1}{x}\right)}^{\sin x}$
Taking ${\log }_e$ to both the sides,
$\Rightarrow \ln L=\underset{x\to 0^{+}}{lim}(-\sin x)\ln x$
$\Rightarrow \ln L=-\underset{x\to 0^{+}}{lim}\frac{\ln x}{\text{cosec}x}\left(\frac{\infty }{\infty }\text{form}\right)$
$\Rightarrow \ln L=-\underset{x\to 0^{+}}{lim}\frac{\frac{1}{x}}{-\text{cosec}x\cdot \cot x}$
$\Rightarrow \ln L=\underset{x\to 0^{+}}{lim}\left(\frac{\sin x}{x}\right)\tan x$
$\Rightarrow \ln L=1\times 0=0\Rightarrow L=1$