Question: Exercise:
---------

> 
> $$\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$$
> 
> 
> 

---

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 $3$.
By applying L'Hospital's rule twice, one has
$$
\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}=\lim\limits\_{x \to 0}{\frac{-\sin x +2 \sin 2x}{\sin x}}=\lim\limits\_{x \to 0}{\frac{-\cos x +4 \cos 2x}{\cos x}}=\frac{-1+4}1=3.
$$