Question

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

Answer 1

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

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

= $\underset{x\to 0}{\text{lim}}\frac{(\text{2 sin}\frac{x}{2}\text{cos}\frac{x}{2}{)}^2}{{\text{sin}}^2\frac{x}{2}}$

= $\underset{x\to 0}{\text{lim}}\frac{{\text{4 sin}}^2\frac{x}{2}{\text{cos}}^2\frac{x}{2}}{{\text{sin}}^2\frac{x}{2}}$

= $\underset{x\to 0}{4}\text{lim}{\text{cos}}^2\frac{x}{2}=4\times 1=4$.