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.
 A: Define
$$F(x)=\frac{1}{x}\log{\left(\frac{\sin{x}}{{x}}\right)}$$
It is now suffice to prove that
$$\lim_{x\to0}F(x)=0$$
Since near $x=0$, we have
$$\log{\left(\frac{\sin{x}}{{x}}\right)}=-\frac{x^2}{6}+O(x^4)$$
So 
$$\lim_{x\to0}F(x)=\lim_{x\to0}\left(-\frac{x}{6}\right)=0$$
A: Hint: First write the expression as

$$e^{\frac{1}{x}\ln(\frac{\sin(x)}{x})} = e^{\frac{1}{x}(-\frac{x^2}{6}+O(x^4))}\longrightarrow_{x\to 0} ...\,. $$

using the Taylor series of $\ln(\frac{\sin(x)}{x})$.
Note:

$$ \ln\left(\frac{\sin(x)}{x}\right) = -\frac{x^2}{6}+O(x^4). $$

A: 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 $\displaystyle\lim_{x\to 0}y=e^0=1$.
A: 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:
$$\lim_{x \to 0}(\frac{\sin x}{x})'=\lim_{x \to 0}\frac{x\cos x-\sin x}{x^2}=^{H}\lim_{x \to 0} \frac{-x\sin x +\cos x-\cos x}{2x}=\lim_{x \to 0}-\frac{1}{2}\sin x=0$$
