Question

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

Answer 1

We have,

$\begin{array}{c}Wehave\\ I={lim}_{x\to 0}\frac{\cos 2x-1}{\cos x-1}\\ ={lim}_{x\to 0}\frac{\left(1-2{\sin }^{2}x\right)-1}{\cos x-1}\left(\therefore \cos 2x=1-{\sin }^{2}x\right)\\ ={lim}_{x\to 0}\frac{-2{\sin }^{2}x}{\cos x-1}\\ ={lim}_{x\to 0}\frac{-2\left(1-{\cos }^{2}x\right)}{\cos x-1}\\ ={lim}_{x\to 0}\frac{-2\left(1-{\cos }^{2}x\right)}{-1\left(1-\cos x\right)}\\ ={lim}_{x\to 0}\frac{2\left(1^2-{\cos }^{2}x\right)}{1-\cos x}\\ ={lim}_{x\to 0}\frac{2\left(1-\cos x\right)\left(1+\cos x\right)}{1-\cos x}\\ ={lim}_{x\to 0}2\left(1+\cos x\right)\\ Puttingx=0\\ =2\left(1+\cos 0\right)\\ =2\left(1+1\right)\\ =4\\ whichistherequiredanswer.\end{array}$