<s>[INST] 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?
 [/INST] 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.
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