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 $3$.
Step 1: find equivalent infinitesimals
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Recall: $\lim\limits\_{x \to 0}{\frac{\cos m - \cos n}{x^2}} = \frac{n^2 - m^2}{2}$. So: $\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{x^2}} = \frac{2^2 - 1^2}{2} = \frac{3}{2}$. To make $RHS = 1$: $\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{\frac{3}{2}x^2}} = 1$. Finally: $\cos x - \cos 2x \sim \frac{3}{2}x^2$.

Recall: $\lim\limits\_{x \to 0}{\frac{1 - \cos x}{x^2}} = \frac{1}{2}$. To make $RHS = 1$: $\lim\limits\_{x \to 0}{\frac{1 - \cos x}{\frac{1}{2}x^2}} = 1$. Finally: $1 - \cos x \sim \frac{1}{2}x^2$

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Step 2: replace with equivalent infinitesimals
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$\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$
$= \lim\limits\_{x \to 0}{\frac{\frac{3}{2}x^2}{\frac{1}{2}x^2}}$
$= 3$

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Answer:
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$$\lim\limits\_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}} = 3$$