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.
 A: 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 
$$\begin{align}\frac{\cos 2x-1}{\cos x-1}&=\frac{\left(1-\frac12 (2x)^2+O(x^4)\right)-1}{\left(1-\frac12 x^2+O(x^4)\right)-1}\\\\&=\frac{-2x^2+O(x^4)}{-\frac12 x^2+O(x^4)}\\\\&=4+O(x^4)\to 4\end{align}$$
A: 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.
A: $\dfrac{1-\cos (2x)}{1-\cos x} = \dfrac{2\sin^2x}{2\sin^2\left(\frac{x}{2}\right)}=4\cdot \left(\dfrac{\sin x}{x}\right)^2\cdot \left(\dfrac{\dfrac{x}{2}}{\sin\left(\dfrac{x}{2}\right)}\right)^2\to 4\cdot 1\cdot 1=4$, as $x \to 0$.
A: Recall $\cos(2x)=2\cos^2(x)-1$ so we may rewrite as
$$\lim\limits_{x\to 0} 2\frac{\cos^2(x)-1}{\cos(x)-1}=\lim\limits_{x\to 0} 2\frac{(\cos(x)-1)(\cos(x)+1)}{\cos(x)-1}=2(\cos(0)+1)=4$$
A: knowing that : 
when x=0 \begin{align*}  1-cos(f(x))\sim \dfrac{1}{2} {f(x)}^{2}
\newline
\newline
  \lim_{x \rightarrow 0}  \dfrac{cos2x-1}{cosx-1} = \lim_{x \rightarrow 0}  \dfrac{-(-cos2x+1)}{-(-cosx+1)} =  \lim_{x \rightarrow 0}  \dfrac{\dfrac{1}{2}{(2x)}^2}{\dfrac{1}{2}{x}^2}=  \lim_{x \rightarrow 0} \dfrac{{(2x)}^2}{{x}^2} =  \lim_{x \rightarrow 0}  \dfrac{4{x}^2}{{x}^2} = 4
\end{align*}
