Q37 / 90 · JEE Main 2012 (26 May Online) · Mathematics

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

  • A
    $-\pi$
  • B
    1
  • C
    $-1$
  • Correct answer
    D
    $\pi$

Solution

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

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

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