Find a limit in an efficent way I'm trying to calculate the following limit:  
$$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$
What I did is writing it as:  
$${e^{\frac{1}{x}\ln \left( {\frac{{\sin x}}{x}} \right)}}$$
Therefore, we need to calculate:  
$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x}$$
Now, we can apply L'Hopital rule, Which I did:
$$\Rightarrow cot(x) - {1 \over x}$$
But in order to reach the final limit two more application of LHR are needed. Is there a better way?
 A: Note that as $x\to 0^+$ you have
$$\frac{\ln\frac{\sin x}x}x\sim \frac{\frac{\sin x}x-1}x\sim \frac{\sin x-x}{x^2}\sim \frac{-x^3}{6x^2}\to 0$$.
A: Elementary proof using well known limits and inequalities:
$$\frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x} =  \frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\frac{{\sin x}}{x}-1 }{x} =\frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\sin x-x }{x^2}$$
but for $x >0$ we have: $x-\frac{x^3}{6} \le \sin x \le x \Rightarrow -\frac{x^3}{6} = x-\frac{x^3}{6}-x\le\sin x -x \le 0 \Rightarrow -\frac{x^2}{6} \le\frac{\sin x-x }{x^2} \le 0$ the rest is squeeze theorem.
A: According to Bernoulli's Inequality, for $0\le x\le1$
$$
\left(\frac{\sin(x)}{x}\right)^{1/x}=\left(1-\frac{x-\sin(x)}{x}\right)^{1/x}\ge1-\frac1x\frac{x-\sin(x)}{x}\tag{1}
$$
This answer provides an elementary proof of Bernoulli's Inequality for rational exponents.
In this answer, it is shown geometrically that for $0\lt x\lt\frac\pi2$, we have $\sin(x)\le x\le\tan(x)$. Therefore,
$$
\begin{align}
0\le\frac{x-\sin(x)}{x^2}&\le\frac{\tan(x)-\sin(x)}{\sin^2(x)}\\[4pt]
&=\frac{\tan(x)-\sin(x)}{\sin(x)\tan(x)}\sec(x)\\[4pt]
&=\frac{1-\cos(x)}{\sin(x)}\sec(x)\\[4pt]
&=\frac{\sin(x)}{1+\cos(x)}\sec(x)\\[4pt]
&\to\frac02\cdot1\\[9pt]
&=0\tag{2}
\end{align}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^2}=0\tag{3}
$$
and therefore,
$$
1\ge\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{1/x}\ge1-0\tag{4}
$$
A: Here, as often, Taylor series are much more efficient than L'Hopital's rule.
$$\begin{eqnarray}\dfrac{\sin(x)}x \ &=&\ \color{#c00}1\color{#0a0}{\,-\,\dfrac{x^2}6+O(x^4)}\!\!\!\!\!\!\!\\&=&\ \color{#c00}1\, +\,\color{#0a0}z\\
\Rightarrow\ \ x^{-1}\log\left(\dfrac{\sin(x)}x\right) &=&\ x^{-1}\log(\color{#c00}1+\color{#0a0}z)\ &=&\  x^{-1}\left(\ \ \ \color{#0a0}z\ \ \ +\ \ O(z^2)\right)\\
&& &=&\ x^{-1}\left(-\color{#0a0}{\dfrac{x^2}6+O(x^4)} \right)\\
&& &=&\ -\dfrac{x}6\, +\ O(x^3)\\
&& &&\!\!\!\!\!\!\!{\rm which}\overset{\phantom{I}} \to \ 0 \ \ \, {\rm as}\, \ \ x\to0^+\\
\end{eqnarray}\qquad\quad\qquad\qquad$$
