Evaluate the Given $operatorname{limit}: lim _{x ightarrow 0} rac{cos 2 x-1}{cos x-1}$

$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$

At $x=0$, the value of the given function takes the form $\frac{0}{0}$.

Now,

$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}=\lim _{x \rightarrow 0} \frac{1-2 \sin ^{2} x-1}{1-2 \sin ^{2} \frac{x}{2}-1} \quad\left[\cos x=1-2 \sin ^{2} \frac{x}{2}\right]$

$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin ^{2} x}{x^{2}}\right) \times x^{2}}{\left(\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right) \times \frac{x^{2}}{4}}$

$=4 \frac{\lim _{x \rightarrow 0}\left(\frac{\sin ^{2} x}{x^{2}}\right)}{\lim _{x \rightarrow 0}\left(\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right)}$

$=4 \frac{\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)^{2}}{\left(\lim _{\frac{x}{2} \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}} \quad\left[x \rightarrow 0 \Rightarrow \frac{x}{2} \rightarrow 0\right]$

$=4 \frac{1^{2}}{1^{2}} \quad\left[\lim _{y \rightarrow 0} \frac{\sin y}{y}=1\right]$

$=4$