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.)
 A: We have $\cos(3x) - \cos(x) = -2\sin(2x)\sin(x)$. Hence,
$$\dfrac{\cos(3x) - \cos(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x) \sin(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x)}{2x} \cdot \dfrac{\sin(x)}{x} \cdot \dfrac{2x^2}{\tan(2x^2)}$$
Now let $x \to 0$ and recall the following limits:
$\lim_{y \to 0}\dfrac{\sin(y)}y = 1$ and $\lim_{y \to 0}\dfrac{\tan(y)}y = 1$.
A: Hints: $\cos 3x = 1 - \dfrac{9x^2}{2} + o(x^2)$, $\cos x = 1 - \dfrac{x^2}{2} + o(x^2)$, and $\tan 2x^2 = 2x^2 + o(x^2)$.
A: Breaking a limit of the form $\frac00$ into $\frac10-\frac10$ is not usually a good way to proceed. Better is to apply L'Hospital twice to the formula as is. Alternately we can use the identity
$$
\cos(x)-\cos(y)=2\sin\left(\frac{y-x}{2}\right)\sin\left(\frac{y+x}{2}\right)
$$
Then
$$
\begin{align}
\frac{\cos(3x)-\cos(x)}{\tan(2x^2)}
&=\frac{-2\sin(x)\sin(2x)}{\tan(2x^2)}\\
&=-2\frac{\sin(x)}{x}\frac{\sin(2x)}{2x}\frac{2x^2}{\tan(2x^2)}\\[6pt]
&\to-2\cdot1\cdot1\cdot1
\end{align}
$$

Using L'Hospital twice
$$
\begin{align}
\lim_{x\to0}\frac{\cos(3x)-\cos(x)}{\tan(2x^2)}
&=\lim_{x\to0}\frac{-3\sin(3x)+\sin(x)}{4x\sec^2(2x^2)}\\
&=\lim_{x\to0}\cos^2(2x^2)\,\lim_{x\to0}\frac{-3\sin(3x)+\sin(x)}{4x}\\
&=1\cdot\lim_{x\to0}\frac{-9\cos(3x)+\cos(x)}{4}\\
&=1\cdot\frac{-8}{4}\\[9pt]
&=-2
\end{align}
$$
