Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}$
$=\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{lim}\frac{1-2{\sin }^{2}x-1}{1-2{\sin }^2\frac{x}{2}-1},$
Using $[\cos x=1-2{\sin }^2\frac{x}{2}]$
$=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{{\sin }^2\frac{x}{2}}=\underset{x\to 0}{lim}\frac{\left(\frac{{\sin }^{2}x}{x^2}\right)\times x^2}{\left(\frac{{\sin }^2\frac{x}{2}}{{\left(\frac{x}{2}\right)}^2}\right)\times \frac{x^2}{4}}$
$=4\cdot \frac{{lim}_{x\to 0}\left(\frac{{\sin }^{2}x}{x^2}\right)}{{lim}_{x\to 0}\left(\frac{{\sin }^2\frac{x}{2}}{{\left(\frac{x}{2}\right)}^2}\right)}=4$