Question

Simplify: $\underset{x\to 0}{lim}\left(\right(sinx{)}^{\frac{1}{x}}+\left(\frac{1}{x}{)}^{sinx}\right)$ for $x>0$

Answer 1

Let ${\left(\sin x\right)}^{\frac{1}{x}}=y$
$\frac{1}{x}\ln \sin x=\ln y$

${lim}_{x\to 0}\frac{\ln \sin x}{x}=-\infty$

$\ln y=-\infty$

$\Rightarrow y=0$

again taking

${\left(\frac{1}{x}\right)}^{\sin x}=z$

$\Rightarrow \sin x\ln \frac{1}{x}=\ln z$

${lim}_{x\to 0}\sin x\times \ln \frac{1}{x}={lim}_{x\to 0}\frac{\ln \frac{1}{x}}{\frac{1}{\sin x}}=0$

$\ln z=0$

$\Rightarrow z=1$

so ${lim}_{x\to 0}({\left(\sin x\right)}^{\frac{1}{x}}+{\left(\frac{1}{x}\right)}^{\sin x})=y+z=1$