Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{1-\cos 2 m x}{1-\cos 2 n x}$
Solution:
$=\lim _{x \rightarrow 0} \frac{1-\cos 2 m x}{1-\cos 2 n x}$
$=\lim _{x \rightarrow 0} \frac{\frac{1-\cos 2 m x}{(2 m x)^{2}} \times(2 m x)^{2}}{\frac{1-\cos 2 n x}{(2 n x)^{2}} \times(2 n x)^{2}}$
$=\frac{\frac{1}{2}}{\frac{1}{2}} \times \frac{\mathrm{m} \times \mathrm{m}}{\mathrm{n} \times \mathrm{n}}\left[\because \frac{1-\cos \theta}{\theta \times \theta}=\frac{1}{2}\right]$
$=\frac{m^{2}}{n^{2}}$
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos 2 m x}{1-\cos 2 n x}=\frac{m^{2}}{n^{2}}$