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

$$\lim_{x \to 0}\frac{\cos 2x-1}{\cos x-1}$$

I have found the above limit using L'Hopital's rule but since this rule is not given in the book so I'm supposed to do it without using this rule.

I know $$\lim_{x \to 0}\frac{1-\cos x}{x}=0$$

I tried to get something of the form of the above limit but I failed to do so.

Kindly help me solve this problem without using L'Hopital's rule.

Answer

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

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