Question

Evaluate the given limit :
$\underset{x\to 0}{lim}\frac{\sin ax}{\sin bx},a,b\ne 0$

Answer 1

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

We need to simplify it,
$\Rightarrow \underset{x\to 0}{lim}\frac{\sin ax}{\sin bx}$
Multiplying and dividing by $ax$ and $bx$
$=(\underset{x\to 0}{lim}\sin ax\times \frac{ax}{ax})÷(\underset{x\to 0}{lim}\sin bx\times \frac{bx}{bx})$
$=\underset{x\to 0}{lim}\frac{\sin ax}{ax}\times \frac{ax}{bx}÷\underset{x\to 0}{lim}\frac{\sin bx}{bx}$
$\{\text{Using}\underset{x\to 0}{lim}\frac{\sin nx}{nx}=1\}$
$=1\times \underset{x\to 0}{lim}\frac{ax}{bx}÷1$
$=\frac{a}{b}$
$\therefore \underset{x\to 0}{lim}\frac{\sin ax}{\sin bx}=\frac{a}{b}$