How do I find $\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan 2x^2}$? Can't understand how to solve limit like this:
$$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}$$
My attempt is:
$$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}=\lim_{x \to 0} \frac{\cos3x}{\tan2x^2}- \frac{\cos x}{\tan2x^2}=\lim_{x \to 0}\frac{1}{\tan2x^2}\left(\cos3x-\cos x \right) = \lim_{x \to 0} \dots$$
And could you explain me how to do next step?
(I know I have to get rid of trigonometry but I don't know how to do this.)

Hints: $\cos 3x = 1 - \dfrac{9x^2}{2} + o(x^2)$, $\cos x = 1 - \dfrac{x^2}{2} + o(x^2)$, and $\tan 2x^2 = 2x^2 + o(x^2)$.
