<s>[INST] Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$ 
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$

Is there a simple way of finding the limit? 
I know the long one: rewrite it as 
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
 [/INST] Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$\cos u=1-\frac{u^2}2+o(u^2),\qquad \sin u=u+o(u)$$
so
\begin{align}
\cos x-\cos 3x&=1-\frac{x^2}2+o(x^2)-\Bigl(1-\frac{9x^2}2+o(x^2)\Bigr)= 4x^2+o(x^2)\\
\sin 3x^2-\sin x^2&=3x^2+o(x^2)-\bigl(\sin x^2+o(x^2)\bigr)=2x^2+o(x^2).
\end{align}
Thus the numerator is equivalent to $4x^2$, the  denominator to $2x^2$, whence
$$\frac{\cos x-\cos 3x}{\sin 3x^2-\sin x^2}\sim_0\frac{4x^2}{2x^2}=2.$$
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