Question

$\underset{x \rightarrow 0}{lim}\frac{tan x \sqrt{tan x} - sin x \sqrt{sin x}}{x^{3} \sqrt{x}}$ equals

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

Answer 1

Answer: (B)

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