Question

Evaluate the given limit :
$\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}$

Answer 1

Given : $\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}$
Substituting $x=0$ in the given limit,
$\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{lim}\frac{\cos 2\left(0\right)-1}{\cos \left(0\right)-1}$
$=\frac{1-1}{1-1}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form.

We need to simplify it,
Let $L=\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}$
$L=\underset{x\to 0}{lim}\frac{(1-2{\sin }^{2}x)-1}{\cos x-1}$
$L=\underset{x\to 0}{lim}\frac{1-2{\sin }^{2}x-1}{\cos x-1}$
$L=\underset{x\to 0}{lim}\frac{-2{\sin }^{2}x}{\cos x-1}$
$L=\underset{x\to 0}{lim}\frac{-2(1-{\cos }^{2}x)}{\cos x-1}$
$L=\underset{x\to 0}{lim}\frac{2(1^2-{\cos }^{2}x)}{1-\cos x}$
$L=\underset{x\to 0}{lim}\frac{2(1-\cos x)(1+\cos x)}{1-\cos x}$
$L=\underset{x\to 0}{lim}2(1+\cos x)\{\because \cos x\ne 1\text{as}x\ne 0\}$
Substituting $x=0$
$L=2(1+\cos 0)$
$\Rightarrow L=2(1+1)$
$\Rightarrow L=2\times 2$
$\Rightarrow L=4$
$\therefore \underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}=4$