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

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}=$
  1. 3
  2. $\frac{3}{2}$
  3. $\frac{3}{4}$
  4. $\frac{3}{16}$

Solution

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

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

Practice more Limits questions on Aicharya