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.

One way of doing it is expanding in Maclaurin series:
$$
\lim_{x \to 0} \frac{1-\cos 2x}{1- \cos 3x} = \lim_{x \to 0} \frac{1-1 +\frac{4x^2}{2} - O(x^4)}{1- 1 + \frac{9x^2}{2} - O(x^4)} = \frac{\frac{4}{2}}{\frac{9}{2}}=\frac{4}{9}
$$
