$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to

$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to
  1. $-\pi$
  2. $\pi$
  3. $\frac{\pi}{2}$
  4. 1

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{\substack{x \rightarrow 0}} \frac{\sin \left(\pi\left\{1-\sin ^2 x\right\}\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2}=\lim _{\substack{2}} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \cdot \frac{\pi \sin ^2 x}{x^2}=\pi\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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