Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{3\sin 2x+2x}{3x+2\tan 3x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{3\sin 2x+2x}{3x+2\tan 3x}$

It becomes 0/0 form

$=\underset{x\to 0}{lim}\frac{2x(\frac{3\sin 2x}{2x}+1)}{3x(1+\frac{2\tan 3x}{3x})}$

$=\underset{x\to 0}{lim}\frac{2(3\frac{\sin 2x}{2x}+1)}{3(1+2\frac{\tan 3x}{3x})}$

$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1]$

$=\frac{2(3\times 1+1)}{3(1+2\times 1)}$

$=\frac{8}{9}$

Therefore,

$\underset{x\to 0}{lim}\frac{3\sin 2x+2x}{3x+2\tan 3x}=\frac{8}{9}$