Question

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

Answer 1

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

Dividing the numerator and the denominator by x:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{\tan 3x}{x}}-2}{3-{\displaystyle \frac{{\sin }^{2}x}{x}}}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{\tan 3x}{3x}}\times 3-2}{3-\left({\displaystyle \frac{\sin x}{x}}\right)\times \sin x}\right] \\ =\frac{3-2}{3-1\times 0}\left[\because \underset{x\to 0}{\lim }\frac{\tan x}{x}=1,\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right] \\ =\frac{1}{3} \end{gathered}$$