Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{\cos c x-1}$
Solution:
$=\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{\cos c x-1}$
$=\lim _{x \rightarrow 0} \frac{1-\cos b x-(1-\cos a x)}{\cos c x-1}$
$=\lim _{x \rightarrow 0} \frac{\frac{(1-\cos b x)}{(b x)^{2}} \times b^{2}-\frac{(1-\cos a x)}{(a x)^{2}} \times a^{2}}{\frac{-(1-\cos c x)}{(c x)^{2}} \times c^{2}}$
$=\frac{a^{2}-b^{2}}{c^{2}}$
$\therefore \lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{\cos c x-1}=\frac{a^{2}-b^{2}}{c^{2}}$