# Evaluate the following limits  $\lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$

$\begin{array}{1 1}4 \\ 1 \\ 0 \\ 2 \end{array}$

At $x = 0$ the value of the given function is $\large\frac{0}{0}.$
We know that: $\quad \cos \: x = 1-2 sin^2 \large\frac{x}{2}$
$\Rightarrow$ $\lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$= \lim\limits_{x \to 0} \large\frac{1-2 \sin^2x-1}{1-2 \sin^2 \large\frac{x}{2} -\normalsize1} = \lim\limits_{x \to 0} \large\frac{ \sin^2x}{ \sin^2 \large\frac{x}{2} } \large\frac{\sin^2x}{\sin^2 \large\frac{x}{2}} =\Large\frac{ \Large\frac{\sin^2x}{x^2} \normalsize \times x^2}{ \Large\frac{\sin^2 \large\frac{x}{2}}{ \large\frac{x}{2} ^2} \normalsize \times \Large\frac{x^2}{4}}$$=$$4 \times \Large\frac { \Large\frac{\sin^2 x}{x^2} } { \Large\frac {\sin^2 \large\frac{x}{2} }{\Large\frac{x^2}{4}} } \Rightarrow \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$ = 4 \large\frac{ \lim\limits_{x \to 0} \bigg( \Large\frac{sin^2x}{x^2} \bigg)}{ \lim\limits_{x \to 0} \bigg( \Large\frac{sin^2 \Large\frac{x}{2}}{ \Large\frac{x}{2}^4}\bigg)}$
As $x \rightarrow 0 \Rightarrow \large\frac{x}{2} $$\rightarrow 0 \Rightarrow \lim\limits_{ x \to 0} \large\frac{\cos\: 2x-1}{\cos\: x - 1}$$ =4 \large\frac{\bigg( \lim\limits_{x \to 0}\Large\frac{sin\: x}{x} \bigg)^2}{ \lim\limits_{ \large\frac{x}{2} \to 0} \bigg( \Large\frac{sin \Large\frac{x}{2}}{ \Large\frac{x}{2} } \bigg)^2}$