Question

$\displaystyle\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^{2} 2 x}=$

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

Answer 1

Answer: (D)

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