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

Notice, $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp \left(\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{\ln(\sin x)-\ln(x)}{x}\right)$$
using L'Hosptal's rule for $\frac 00$ form, 
$$=\lim_{x\to 0}\exp \left(\frac{\frac{\cos x}{\sin x}-\frac1x}{1}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{x\cos x-\sin x}{x\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{x(-\sin x)+\cos x-\cos x}{x\cos x+\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\sin x}{x\cos x+\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\cos x-\sin x}{-x\sin x+\cos x+\cos x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\cos x-\sin x}{-x\sin x+2\cos x}\right)$$
$$=\exp \left(\frac{0}{0+2}\right)=e^0=\color{red}{1}$$
