limit of $f(x) = \lim \limits_{x \to 0} (\frac{\sin x}{x})^{\frac 1x}$ Any ideas how to calculate this limit without using taylor?
 $$f(x) = \lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac1x}$$

With Taylor expansions: (just for reference)
We will use 
$$\begin{align}
\sin u &= u + o(u^2) \\
\ln(1+x) &= u  + o(u)
\end{align}$$
when $u\to0$. (In particular, $\ln(1+o(u)) = o(u)$.)

Write
$$
\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}
= \left(\frac{x+o(x^2)}{x}\right)^{\frac{1}{x}}
= \left(1+o(x)\right)^{\frac{1}{x}}
= e^{ \left(\frac{1}{x}\ln\left(1+o(x)\right)\right) }
= e^{ \left(\frac{1}{x}\left(o(x)\right)\right) }
= e^{o(1) }
$$
so the limit is $e^0=1$.
