TITLE: Finding the limit of $\frac{1-\cos(2x)}{1-\cos(3x)}$ for $x \to 0$
QUESTION [6 upvotes]: As $x$ goes to $0$, what is the limit of
$$\frac{1-\cos(2x)}{1-\cos(3x)}$$
Thanks.

REPLY [5 votes]: Putting $x=2y,$
$$\lim_{x \rightarrow 0}\frac{1-\cos(2x)}{1-\cos(3x)}$$
$$=\lim_{y \rightarrow 0}\frac{1-\cos4y}{1-\cos6y}$$
$$=\lim_{y \rightarrow 0}\frac{2\sin^22y}{2\sin^23y} \text{ as }\cos2\theta=1-2\sin^2\theta$$
$$=\lim_{y \rightarrow 0} \frac{4\cdot\left(\frac{\sin 2y}{2y}\right)^2}{9\cdot \left(\frac{\sin 3y}{3y}\right)^2}$$
$$=\frac49\text{ as  }\lim_{h\to0}\frac{\sin h}h=1$$

Alternatively, 
$$\frac{1-\cos2x}{1-\cos3x}=\left(\frac{\sin2x}{\sin3x}\right)^2\cdot\left(\frac{1+\cos3x}{1+\cos2x}\right)$$
Now, $\lim_{x\to0}\cos ax=1$ for finite $a$
$$\implies \lim_{x\to0}\left(\frac{1+\cos3x}{1+\cos2x}\right)=\frac{1+1}{1+1}=1$$
and $$\lim_{x\to0}\left(\frac{\sin2x}{\sin3x}\right)=\frac23\cdot \frac{\lim_{x\to0}\frac{\sin2x}{2x}}{\lim_{x\to0}\frac{\sin3x}{3x}}=\frac23$$