Q132 / 200 · AP EAMCET 2001 · Mathematics

$\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$ is equal to

  • A
    $0$
  • Correct answer
    B
    $1$
  • C
    $-1$
  • D
    $\infty$

Solution

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

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

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