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: 
By applying L'Hospital's rule twice, one has
$$
\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}=\lim\limits\_{x \to 0}{\frac{-\sin x +2 \sin 2x}{\sin x}}=\lim\limits\_{x \to 0}{\frac{-\cos x +4 \cos 2x}{\cos x}}=\frac{-1+4}1=3.
$$
If we evaluate directly we get the fraction $\frac{0}{0}$. So using L'Hopital's rule:
$$
\lim\_{x \rightarrow 0} \frac{\cos x - \cos 2x}{1-\cos x}
=
\lim\_{x \rightarrow 0} \frac{-\sin x + 2\sin 2x}{\sin x}
$$

Now, if we evaluate directly once more, we get the fraction $\frac{0}{0}$. So we use L'Hopital again to get:

$$
\lim\_{x \rightarrow 0} \frac{-\cos x + 4 \cos 2x}{\cos x}
$$

Which, when evaluated on $x=0$ yields: $$\frac{-1 + 4}{1} = 3$$