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

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

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 $\frac{\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x^2}{2}\right)}{\left(\frac{x^2}{4}\right)}\right]}{\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \times \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{x^2}{2}\right)}{\left(\frac{x}{2}\right)}}=\frac{\left.x^2\right) \times \frac{1}{2}}{\left[\frac{x}{2}\right)}=\frac{1}{\sqrt{2}} \times \frac{2}{1}=\sqrt{2}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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