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 $2$.

$$
\cos x-\cos(3x) = \cos x\left(\sin^2x+3\sin^2 x\right)
$$
then
$$
\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \frac{x^2}{x^2}\cos x\left(\frac{\sin^2x+3\sin^2 x}{\sin (3x^2)-\sin (x^2)}\right)=\cos x\left(\frac{\left(\frac{\sin x}{x}\right)^2+3\left(\frac{\sin x}{x}\right)^2}{\frac{3\sin(3x^2)}{3x^2}-\frac{\sin (x^2)}{x^2}}\right)
$$
hence
$$
\lim_{x\to 0}=\cos x\left(\frac{\left(\frac{\sin x}{x}\right)^2+3\left(\frac{\sin x}{x}\right)^2}{\frac{3\sin(3x^2)}{3x^2}-\frac{\sin (x^2)}{x^2}}\right) = 1\cdot\left(\frac{1+3}{3-1}\right) = 2
$$
