Explanation

(c)
Here, $\lim_{x \rightarrow 0}{\left( \sin x \right)^{1/x} + \lim_{x \rightarrow 0}\left( \frac{1}{x} \right)^{\sin x}} = 0 + \lim_{x \rightarrow 0}e^{\log\left( \frac{1}{x} \right)^{\sin x}}$
$\left\lbrack \begin{array}{r}
\lim_{x \rightarrow 0}\ \left( \sin x \right)^{\frac{1}{x}} \rightarrow 0 \\
as,\ \ \ \ \ 0 < \sin{x < 1}
\end{array} \right\rbrack$
$= e^{\lim_{x \rightarrow 0}\frac{\log{(1/x)}}{cosec\ x}} = e^{\lim_{x \rightarrow 0}\frac{x\left( - \frac{1}{x^{2}} \right)}{- cosec\ x\cot x}}$
[by L’Hospital’s rule]
$= e^{\lim_{x \rightarrow 0}\frac{\sin x}{x}\tan x} = e^{0} = 1$