Answer
$\dfrac{1}{\sqrt[6] e}$
Work Step by Step
Write the Taylor series for the function as: $\sin x=x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+.......=\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{(2k+1)!}$
Now, we have:
$\lim\limits_{x \to 0} [\dfrac{\sin (x)}{x}]^{1/x^2}=\lim\limits_{x \to 0} [\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^{2k}}{(2k+1)!}]^{1/x^2}\\=\lim\limits_{x \to 0} [(-1)^0 \dfrac{x^{2(0)+1}}{(2(0)+1)!}+(-1)^{1} \dfrac{x^{2(1)+1}}{(2(1)+1)!}+.......]^{1/x^2}\\=\lim\limits_{x \to 0} (1-\dfrac{x^2}{6}+\dfrac{x^4}{5!}-......)^{1/x^2}\\=e^{-1/6}\\=\dfrac{1}{\sqrt[6] e}$
Therefore, we have: $\lim\limits_{x \to 0} [\dfrac{\sin (x)}{x}]^{1/x^2}=\dfrac{1}{\sqrt[6] e}$
