Question

Evaluate :
$i.$ $\underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}$
$ii.$ $\underset{x\to 0}{lim}\frac{\tan x}{x}$

Answer 1

$\left(i\right)$ Given : $\underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}$
Substituting the given value,
$\underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}=\frac{\sin 4\left(0\right)}{\sin 2\left(0\right)}$
$=\frac{\sin 0}{\sin 0}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form, we need to simplify it.

$=\left(\underset{x\to 0}{lim}\sin 4x\right)÷\left(\underset{x\to 0}{lim}\sin 2x\right)$
$=(\underset{x\to 0}{lim}\sin 4x\cdot \frac{4x}{4x})÷(\underset{x\to 0}{lim}\sin 2x\cdot \frac{2x}{2x})$

$=(\underset{x\to 0}{lim}\frac{\sin 4x}{4x}\times \underset{x\to 0}{lim}\frac{4x}{2x})÷\left(\underset{x\to 0}{lim}\frac{\sin 2x}{2x}\right)$

We know that $\underset{x\to 0}{lim}\frac{\sin nx}{nx}=1$
$=1\times \underset{x\to 0}{lim}2\times 1=1\times 2\times 1$
$=2$
$\therefore \underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}=2$

$\left(ii\right)$ Given: $\underset{x\to 0}{lim}\frac{\tan x}{x}$
$=\underset{x\to 0}{lim}\frac{1}{x}\times \tan x$
$=\underset{x\to 0}{lim}\frac{1}{x}\times \frac{\sin x}{\cos x}$
$=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \frac{1}{\cos x}$
$=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{x\to 0}{lim}\frac{1}{\cos x}$

We know that $\underset{x\to 0}{lim}\frac{\sin nx}{nx}=1$
$=1\times \frac{1}{\cos 0}$
$=1$
$\therefore \underset{x\to 0}{lim}\frac{\tan x}{x}=1$