Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.

Write
\begin{eqnarray*}
\frac{2\sin x-\sin 2x}{x-\sin x} &=&\frac{2\sin x-2\sin x\cos x}{x-\sin x} \\
&=&2\left( \frac{\sin x}{x}\right) \left( \frac{1-\cos x}{x^{2}}\right)
\left( \frac{x^{3}}{x-\sin x}\right) 
\end{eqnarray*}
Using standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6}
\end{eqnarray*}
it follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x-\sin x} &=&2\lim_{x\rightarrow
0}\left( \frac{\sin x}{x}\right) \lim_{x\rightarrow 0}\left( \frac{1-\cos x}{%
x^{2}}\right) \lim_{x\rightarrow 0}\left( \frac{x^{3}}{x-\sin x}\right)  \\
&=&2\left( 1\right) \left( \frac{1}{2}\right) \left( \frac{6}{1}\right)  \\
&=&6.
\end{eqnarray*}
