Question

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

• A. 3
• B. $\frac{3}{2}$
• C. $\frac{3}{4}$
• D. $\frac{3}{16}$

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{{\tan }^{2}2x}(\frac{0}{0}$ form $)$
On rationalising
$=\underset{x\to 0}{\lim }\frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{{\tan }^{2}2x}\times \frac{\sqrt{1+x\sin x}+\sqrt{\cos x}}{\sqrt{1+x\sin x}+\sqrt{\cos x}}$
$=\underset{x\to 0}{\lim }\frac{1+x\sin x-\cos x}{{\tan }^{2}2x}\times \underset{x\to 0}{\lim }\frac{1}{\sqrt{1+x\sin x}+\sqrt{\cos x}}$
$=\underset{x\to 0}{\lim }\frac{(1-\cos x)+x\sin x}{{\tan }^{2}2x}\times \frac{1}{2}$
$=\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{x}{2}+2x\sin \frac{x}{2}\cos \frac{x}{2}}{{\tan }^{2}2x}\times \frac{1}{2}$
$=\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{x}{2}\left[1+\frac{x}{\tan \frac{x}{2}}\right]}{{\tan }^{2}2x}\times \frac{1}{2}$
$=\underset{x\to 0}{\lim }\frac{\frac{1}{4}\cdot {\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2\left[1+\frac{2\cdot \frac{x}{2}}{\tan \frac{x}{2}}\right]}{4{\left[\frac{\tan 2x}{2x}\right]}^2}$
On applying limits, we get
$=\frac{\frac{1}{4}\times (1{)}^2\times [1+2\times 1]}{4(1{)}^2}=\frac{3}{16}$