Help Evaluating $\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$ Does anyone know how to evaluate the following limit?
$\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$
The answer is 1 , but I want to see a step by step solution if possible.

Let $\displaystyle y=(\frac{\sin x}{x})^{\frac{1}{x}}$.  Then $\displaystyle\lim_{x\to 0} \ln y=\lim_{x\to 0}\ln \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{\ln\sin x-\ln x}{x}=^{(LH)}\lim_{x\to 0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{1}$
$\displaystyle=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\sin x+\cos x-\cos x}{x\cos x+\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\cos x-\sin x}{-x\sin x+\cos x+\cos x}=\frac{0}{2}=0,$
so $\displaystyle\lim_{x\to 0}y=e^0=1$.
