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.

Instead of using power series, which feels like a way of hiding the use of derivatives and, ultimately, L'Hôpital, we can use a trigonometric identity and take limits.
Since
$$
\frac{4\sin(x)-2\sin(2x)}{2\sin(2x)-\sin(4x)}
=\frac1{2(1+\cos(x))\cos(x)}\tag{1}
$$
we have
$$
\lim_{x\to0}\frac{2^{k+2}\sin\left(\frac{x}{2^{k+2}}\right)-2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)}{2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)-2^k\sin\left(\frac{x}{2^k}\right)}
=\frac14\tag{2}
$$
Taking the product of $(2)$ from $k=0$ to $k=n-1$, we get
$$
\lim_{x\to0}\frac{2^{n+1}\sin\left(\frac{x}{2^{n+1}}\right)-2^n\sin\left(\frac{x}{2^n}\right)}{2\sin\left(\frac x2\right)-\sin\left(x\right)}
=\frac1{4^n}\tag{3}
$$
Summing $(3)$ from $n=0$ to $n=\infty$, we get
$$
\lim_{x\to0}\frac{x-\sin(x)}{2\sin\left(\frac x2\right)-\sin\left(x\right)}=\frac43\tag{4}
$$
Applying $(1)$ gives
$$
\begin{align}
\lim_{x\to0}\frac{2\sin\left(\frac x2\right)-\sin\left(x\right)}{2\sin(x)-\sin(2x)}
&=\lim_{x\to0}\frac12\frac{4\sin\left(\frac x2\right)-2\sin\left(x\right)}{2\sin(x)-\sin(2x)}\\
&=\frac18\tag{5}
\end{align}
$$
Multiplying $(4)$ and $(5)$, and taking the reciprocal, yields
$$
\begin{align}
\lim_{x\to0}\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}=6\tag{6}
\end{align}
$$
