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

Question

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

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

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

Answer 2

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.

Answer 3

$\dfrac{1-\cos (2x)}{1-\cos x} = \dfrac{2\sin^2x}{2\sin^2\left(\frac{x}{2}\right)}=4\cdot \left(\dfrac{\sin x}{x}\right)^2\cdot \left(\dfrac{\dfrac{x}{2}}{\sin\left(\dfrac{x}{2}\right)}\right)^2\to 4\cdot 1\cdot 1=4$, as $x \to 0$.

Answer 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

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