Question

The value of $\underset{x\to 0}{lim}{\left(\frac{1}{x}\right)}^{\sin x}$, where $x>0$ is

Answer 1

Let $L=\underset{x\to 0}{lim}{\left(\frac{1}{x}\right)}^{\sin x}$
Taking $\ln$ both sides,
$\Rightarrow \ln L=\underset{x\to 0}{lim}\sin x\ln \left(\frac{1}{x}\right)$
$\Rightarrow$ $L=e^{\underset{x\to 0}{lim}\frac{-\ln x}{\text{cosec}x}}$
Using L Hospital's rule
$=e^{\underset{x\to 0}{lim}\frac{-\frac{1}{x}}{-\text{cosec}x\cot x}}$
$=e^{\underset{x\to 0}{lim}\frac{\sin x}{x}\tan x}=e^0=1$