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.

First write limit in the other form:
$$\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}=\lim_{x \to 0}e^{\ln\frac{\sin{x}}{{x}}}=\lim_{x \to 0}e^{\frac{1}{x}\ln{\frac{\sin x}{x}}}=e^{\lim_{x \to 0}\frac{1}{x}\ln{\frac{sin}{x}}}$$
Now evaluate limit $\displaystyle \lim_{x \to 0}\frac{1}{x}\ln{(\frac{sin}{x})^{\frac{1}{x}}}$. You can use l'Hospital rule.
$$\lim_{x \to 0}\frac{1}{x}\ln{\frac{sin}{x}}=\lim_{x \to 0}\frac{(\ln\frac{sin}{x})'}{x'}=\lim_{x \to 0} (\frac{\sin x}{x})'\cdot \frac{x}{\sin x} =\lim_{x \to 0}\frac{x}{\sin x}\cdot \lim_{x \to 0}(\frac{\sin x}{x})'$$
Because $\displaystyle \lim_{x \to 0}\frac{x}{\sin x} = 1 $ you have to show that $\displaystyle \lim_{x \to 0}(\frac{\sin x}{x})'=0$. Use l'Hospital rule again:
$$\lim_{x \to 0}(\frac{\sin x}{x})'=\lim_{x \to 0}\frac{x\cos x-\sin x}{x^2}=^{H}\lim_{x \to 0} \frac{-x\sin x +\cos x-\cos x}{2x}=\lim_{x \to 0}-\frac{1}{2}\sin x=0$$
