Find $\lim_{x \to 0}\frac{\cos 2x-1}{\cos x-1}$ without L'Hopital's rule.

Recall $\cos(2x)=2\cos^2(x)-1$ so we may rewrite as

$$\lim\limits_{x\to 0} 2\frac{\cos^2(x)-1}{\cos(x)-1}=\lim\limits_{x\to 0} 2\frac{(\cos(x)-1)(\cos(x)+1)}{\cos(x)-1}=2(\cos(0)+1)=4$$

Since others have already presented a great way forward, I thought that it would be instructive to present an alternative approach. To that end, we use

$$\cos x=1-\frac12 x^2+O(x^4)$$

to write

$$\begin{align}\frac{\cos 2x-1}{\cos x-1}&=\frac{\left(1-\frac12 (2x)^2+O(x^4)\right)-1}{\left(1-\frac12 x^2+O(x^4)\right)-1}\\\\&=\frac{-2x^2+O(x^4)}{-\frac12 x^2+O(x^4)}\\\\&=4+O(x^4)\to 4\end{align}$$

Hint: try to expand $\cos 2x$ first.

Remember $\cos 2x=\cos^2x-\sin^2x$, or $\cos(2x)=2\cos^2x-1$. You can continue solve this problem by using this identity.