#### Find the limit of the following:$\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}$Option: 1 $1$Option: 2 $0$Option: 3 $\infty$Option: 4 $e$

Given: $\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}$

If $\lim _{x \rightarrow a} f(x)=1$ and $\lim _{x \rightarrow a} g(x)=\infty$, then $\lim _{x \rightarrow a} f(x)^{g\left ( x \right )}= e^{\lim _{x \rightarrow a} \left ( f\left ( x \right ) -1\right )\left ( g\left ( x \right ) \right )}$.

$=\lim _{x \rightarrow 0}\left[1+\left(\frac{\sin x}{x}-1\right)\right]^{1 / x}$
$=e^{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-1\right) \frac{1}{x}}$
$=e^{\lim _{x \rightarrow 0}\left(\frac{\sin x-x}{x^{2}}\right) }$

Applying L-Hospital’s rule, the above expression becomes,

$=e^{\lim _{x \rightarrow 0}\left(\frac{\cos x-1}{2 x}\right)}$
$=e^{\frac{-1}{2} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x}\right)}$

Using the property of limit $\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=0$,then the expression becomes,

$=e^0$
$=1$

Therefore,$\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}= 1$