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 $2$.

Use the Maclaurin series
$$\cos x  = 1-1/2 \; x^2+1/24 \; x^4 +O(x^6)$$
$$\cos 3x = 1-9/2 \; x^2+27/8 \; x^4 +O(x^6)$$ 
$$\sin x^2 = x^2-1/6\;  x^6+O(x^8)$$
$$\sin 3x^2 = 3x^2-9/2 \; x^6+O(x^8)$$
then quotient is 
$$\frac{-1/2 + 9/2}{3-1} + O(x^2)$$
and therefore the limit is 2.
