The value of \(\lim _{x \rightarrow 0}\left((\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)\), where \(x\gt0\) is

\(\begin{aligned} & \lim _{x \rightarrow 0^0}\left\{(\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right\} \\ & =\lim _{x \rightarrow 0}(\sin x)^{\frac{1}{x}}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\ & =0+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\ & =\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}\end{aligned}\)
Let \(l=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}\). Then,
\(\begin{aligned}
& \log l=\log \lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\
& \Rightarrow \log l=\lim _{x \rightarrow 0}(-\sin x \log x) \\
& \Rightarrow \log l=-\lim _{x \rightarrow 0} \frac{\log x}{\operatorname{cosec} x} \\
& \Rightarrow \log l=-\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{-\operatorname{cosec} x \cot x}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x \cos x} \\
& \Rightarrow \log l=\lim _{x \rightarrow 0} \frac{\tan x}{x} \times \sin x=1 \times 0=0 \\
& \Rightarrow l=\mathrm{e}^0=1
\end{aligned}\)