Question

Evaluate the given limit :
$\underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}$

Answer 1

Given : $\underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}$
Substituting $x=0$ in the given limit,
$\underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}=\underset{x\to 0}{lim}\frac{a\left(0\right)+\left(0\right)\cos 0}{b\sin 0}$
$=\frac{0+0}{0}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form.

We need to simplify it,
Let $L=\underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}$
$L=\underset{x\to 0}{lim}\frac{x(a+\cos x)}{b\sin x}$
$L=\underset{x\to 0}{lim}(\frac{a+\cos x}{b}\times \frac{x}{\sin x})$
$L=\underset{x\to 0}{lim}(\frac{(a+\cos x)}{b}÷\frac{\sin x}{x})$
$L=\left(\underset{x\to 0}{lim}\frac{a+\cos x}{b}\right)÷\left(\underset{x\to 0}{lim}\frac{\sin x}{x}\right)$
$L=\underset{x\to 0}{lim}\left(\frac{a+\cos x}{b}\right)÷1$
$L=\underset{x\to 0}{lim}\left(\frac{a+\cos x}{b}\right)$
Substituting $x=0$
$L=\frac{a+\cos 0}{b}$
$L=\frac{a+1}{b}$
$\therefore \underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}=\frac{a+1}{b}$