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?
 $$f(x) = \lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac1x}$$

Take the log of both sides and examining the 2 sides of the limits yields
$$\ln L^+=\lim \limits_{x \to 0^+} \frac{\ln (\sin x)-\ln(x)}{x}$$
$$\ln L^-=\lim \limits_{x \to 0^-} \frac{\ln (\sin (-x))-\ln(-x)}{x}$$
which can be solved by L'Hopital's to both equal $0$, so the limit is 1.
