Question

Evaluate the limit: $\underset{x\to 0}{lim}\frac{\tan 3x-2x}{3x-{\sin }^{2}x}$

Answer 1

We have,

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

It becomes 0/0 form.

Taking $x$ common from numerator and denominator.

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

$=\underset{x\to 0}{lim}\frac{(\frac{\tan 3x}{x}-2)}{(3-\frac{\sin x}{x}(\sin x\left)\right)}$

$=\underset{x\to 0}{lim}\frac{3\times \frac{\tan 3x}{3x}-2}{(3-\frac{\sin x}{x}(\sin x\left)\right)}$

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

$=\frac{3-2}{3-1\left(0\right)}$

$=\frac{1}{3}$

Therefore,

$\underset{x\to 0}{lim}\frac{\tan 3x-2x}{3x-{\sin }^{2}x}=\frac{1}{3}$