Question

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

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (B)

Let, $L={\lim }_{x\to 0}\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt{x}}$
$={\lim }_{x\to 0}\frac{(\tan x{)}^{\frac{3}{2}}\left[1-(\cos x{)}^{\frac{3}{2}}\right]}{x^{\frac{3}{2}}\cdot x^2}$
$=1^{\frac{3}{2}}\cdot {\lim }_{x\to 0}\frac{1-{\cos }^{3}x}{x^2}\cdot \frac{1}{1+(\cos x{)}^{\frac{3}{2}}}$ (Rationalizing)
$={\lim }_{x\to 0}\frac{1-\cos x}{x^2}\cdot \left(1+\cos x+{\cos }^{2}x\right)\cdot \frac{1}{1+(\cos x{)}^{\frac{3}{2}}}$
$=\frac{1}{2}\cdot \frac{1}{2}(1+1+1)=\frac{3}{4}$