Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}$
Solution:
$=\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}$
$=\lim _{x \rightarrow 0} \frac{a x}{b \sin x}+\frac{x \cos x}{b \sin x}$
$=\lim _{x \rightarrow 0} \frac{a x}{b \sin x}+\lim _{x \rightarrow 0} \frac{x \cos x}{b \sin x}$
$\left.=\frac{a}{b}+\frac{1}{b}\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]_{x \rightarrow 0} \cos \theta=1\right]$
$=\frac{a+1}{b}$
$\therefore \lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}=\frac{a+1}{b}$