Question: Exercise:
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> $$\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$$
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I've posted my solution down below, however if there are more elegant approaches, feel free to include your own solutions.

Answer: 
We have the double angle formula:

$$\cos(2x)=2\cos^2(x)-1$$

Thus, it follows that

$$\cos(x)-\cos(2x)=1+\cos(x)-2\cos^2(x)=(1-\cos(x))(2\cos(x)+1)$$

Thus, we find

$$\frac{\cos(x)-\cos(2x)}{1-\cos(x)}=2\cos(x)+1$$

So the limit is directly $3$.
$=\dfrac{\cos x - \cos 2x}{1-\cos x}$

$=\dfrac{2\sin \left( \frac{x+2x}{2}\right) \sin \left( \frac{2x-x}{2}\right) }{2\sin^2 \left( \frac{x}{2}\right)}$

$=\dfrac{\sin 3x/2}{x} . \dfrac{x}{\sin x/2}$

$=\left(3/2\right) . \frac{1}{\left( 1/2\right)}$

$=3$

Formulae involved--

$\cos A - \cos B = 2\sin \frac{A+B}{2} \sin \frac{B-A}{2}$

$\cos 2A = 1-2\sin^2 A$

$\lim\limits\_{x\to 0} \dfrac {\sin ax}{x} =a$