Q99 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

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

  • A
    $-\pi$
  • Correct answer
    B
    $\pi$
  • C
    $\frac{\pi}{2}$
  • D
    1

Solution

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

Hence, the correct answer is option: 'B'.

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