Question

$\underset{x\to 0}{lim}\frac{tanx\sqrt{tanx}-sinx\sqrt{sinx}}{x^3\sqrt{x}}$ equals

• A. 1/4
• B. 3/4
• C. 1/2
• D. 1

Answer 1

The correct option is B 3/4

$\begin{array}{c}Given,\\ {lim}_{x\to 0}\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt{x}}\\ Now,\\ \Rightarrow {lim}_{x\to 0}\frac{{\left(\tan x\right)}^{\frac{3}{2}}-{\left(\sin x\right)}^{\frac{3}{2}}}{x^3\sqrt{x}\left[{\left(\tan x\right)}^{\frac{3}{2}}-{\left(\sin x\right)}^{\frac{3}{2}}\right]}\\ \Rightarrow {lim}_{x\to 0}\frac{{\tan }^{3}x-{\sin }^{3}x}{x^{\frac{7}{2}}{x}^{\frac{3}{2}}\left[{\left(\frac{\tan x}{x}\right)}^{\frac{3}{2}}-{\left(\frac{\sin x}{x}\right)}^{\frac{3}{2}}\right]}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{\left(\tan x-\sin x\right)\left({\tan }^{2}x+{\sin }^{2}x+\tan x\sin x\right)}{x^5}\\ =\frac{1}{2}{lim}_{x\to 0}\left(\frac{\tan x-\sin x}{x^3}\right)\left(\frac{{\tan }^{2}x}{x^2}+\frac{{\sin }^{2}x}{x^2}+\frac{\tan x}{x}\frac{\sin x}{x}\right)\\ =\frac{3}{2}{lim}_{x\to 0}\frac{x+\frac{x^3}{3}+........\left(x-\frac{x^3}{6}\tan x\right)}{x^3}\\ =\left(\frac{1}{3}+\frac{1}{6}\right)\times \frac{3}{2}\\ =\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}\end{array}$

Option $B$ is correct answer.