Finding the limit of $\frac{1-\cos(2x)}{1-\cos(3x)}$ for $x \to 0$ As $x$ goes to $0$, what is the limit of
$$\frac{1-\cos(2x)}{1-\cos(3x)}$$
Thanks.
 A: 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 $$\lim_{x\to0}\left(\frac{\sin2x}{\sin3x}\right)=\frac23\cdot \frac{\lim_{x\to0}\frac{\sin2x}{2x}}{\lim_{x\to0}\frac{\sin3x}{3x}}=\frac23$$
A: One way to continue with your idea is to notice that
$$\lim_{x \to 0} \frac{2-\color{red}2 cos(x)^2}{1+3cos(x)-4cos(x)^3} $$
is equal to
$$\lim_{y \to 1^-} \frac{2- \color{red}2  y^2}{1+3y-4y^3} $$
after making the change of variable $y = \cos x$. (With practice, you could learn to use the ideas this suggests without making the change of variable!)
(edit: red numerals to indicate corrections from the original expression)
A: Use the L'Hopital rule.
$$\lim_{x \rightarrow 0}\frac{1-\cos(2x)}{1-\cos(3x)}=\lim_{x \rightarrow 0}\frac{2\sin(2x)}{3\sin(3x)}=\lim_{x \rightarrow 0}\frac{4x}{9x}=\frac49$$
A: One way of doing it is expanding in Maclaurin series:
$$
\lim_{x \to 0} \frac{1-\cos 2x}{1- \cos 3x} = \lim_{x \to 0} \frac{1-1 +\frac{4x^2}{2} - O(x^4)}{1- 1 + \frac{9x^2}{2} - O(x^4)} = \frac{\frac{4}{2}}{\frac{9}{2}}=\frac{4}{9}
$$
