Question

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

Answer 1

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

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

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

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

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

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

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

$={lim}_{x\to 0}2(1+\cos x)$

Putting $x=0$

$=(1+\cos 0)$

$=2(1+1)$

$=2\times 2$

$=4.$

Hence, the answer is $4.$