$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
I tried the following but it doesn't seem to work...
$$= \lim_{x\to 0} \frac{x}{2} \cdot \frac{\sin(3x)}{3x}\cdot\frac{6x}{1-\cos(6x)}$$
$$= 0$$
But the result of this limit is $\frac{1}{6}$.
Am I missing something or did I make any glaring mistakes?

More generally:
All you need is
$\dfrac{\sin(x)}{x} \to 1$
and
$1-\cos(x)
=2\sin^2(x/2)
$.
$\begin{array}\\
\dfrac{x\sin(ax)}{1-\cos(bx)}
&=ax^2\dfrac{\sin(ax)}{ax}\dfrac{1}{2\sin^2(bx/2)}\\
&=\dfrac{ax^2}{2(b/2)^2}\dfrac{\sin(ax)}{ax}\dfrac{(b/2)^2}{\sin^2(bx/2)}\\
&=\dfrac{a}{b^2/2}\dfrac{\sin(ax)}{ax}\dfrac{(bx/2)^2}{\sin^2(bx/2)}\\
&\to \dfrac{2a}{b^2}\\
\end{array}
$
For $a=3, b=6$
this is
$\dfrac{2\cdot 3}{6^2}
=\dfrac16
$.
