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.
 A: 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
$$
\begin{align}
\lim_{x\to0}\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}=6\tag{6}
\end{align}
$$
A: 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*}
A: We have
$$\sin x=x-\frac16x^3+O(x^5)\implies 2\sin x-\sin 2x =x^3+O(x^5)$$
and 
$$x-\sin x=\frac16x^3+O(x^5)$$
Thus
$$\frac{2\sin x-\sin 2x}{x-\sin x}=6+O(x^2)\to 6\,\,\text{as}\,\,x\to 0$$
A: Just as Corindo and Dr. MV answered, for this kind of problems, Taylor expansions are very useful not only for obtaing the limit but also to detect how it is approached.
Using Taylor series for $$\sin(y)=y-\frac{y^3}{6}+\frac{y^5}{120}+O\left(y^6\right)$$ and replacing $y$ by $2x$ in numerator and $x$ in denominator, we then have $$A=\frac{2\sin x-\sin 2x}{x-\sin x}=\frac{2(x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots)-(2 x-\frac{4 x^3}{3}+\frac{4 x^5}{15}+\cdots)}{x-(x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots)}$$ $$A=\frac{x^3-\frac{x^5}{4}+\cdots}{\frac{x^3}{6}-\frac{x^5}{120}+\cdots}$$ Performing the long division, we then have $$A=6-\frac{6 x^2}{5}+\cdots$$
If you plot on the same graph the original function and this last approximation for the range $-1\leq x \leq 1$, you will probably be amazed to see how close the two curves are.
A: Use this 2 relations:
$$\sin(x) = x - \frac{x^3}{6} + o(x^4) = x + o(x^2)$$
$$ \cos(x)  = 1 - \frac{x^2}{2} + o(x^3)$$
