<s>[INST] $\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
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?
 [/INST] We can also make the substitution $x \mapsto \frac{u}{3}$:
$$\frac{1}{3} \frac{u \sin u}{1 - \cos 2u} =  \frac{1}{3} \frac{u \sin u}{1 - \cos^2 u + \sin^2 u} = \frac{1}{3} \frac{u \sin u}{2 \sin^2 u} = \frac{1}{6}\frac{u}{\sin u}.$$
and since $\frac{u}{\sin u} \to 1$ as $u \to 0$:
$$\lim_{u \to 0} \frac{1}{3} \frac{u \sin u}{1- \cos 2u} = \frac{1}{6} \times 1 = \frac{1}{6}.$$
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