Q76 / 150 · MHT CET 2021 (24 Sep Shift 2) · Mathematics
$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x}=$
- Correct answerA$\sqrt{2}$
- B$\frac{1}{\sqrt{2}}$
- C0
- D$\frac{1}{2}$
Solution
$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 \frac{x^2}{2}}}{2 \sin ^2 \frac{x}{2}}=\lim _{x \rightarrow 0} \frac{\sqrt{2} \sin \frac{x^2}{2}}{2 \sin ^2 \frac{x}{2}}
\end{aligned}$
Dividing numerator and denominator by $\frac{x^2}{4}$, we get
$\frac{\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x^2}{2}\right)}{\left(\frac{x^2}{4}\right)}\right]}{\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \times \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{x^2}{2}\right)}{\left(\frac{x}{2}\right)}}=\frac{\left.x^2\right) \times \frac{1}{2}}{\left[\frac{x}{2}\right)}=\frac{1}{\sqrt{2}} \times \frac{2}{1}=\sqrt{2}$
Hence, the correct answer is option: 'A'.