Q99 / 180 · COMEDK 2020 · Mathematics

$\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ is equal to

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

Solution

We have, $$ \begin{aligned} \lim _{x \rightarrow 1} & \frac{\tan \left(x^{2}-1\right)}{x-1} \\ &=\lim _{x \rightarrow 1} \frac{\sec ^{2}\left(x^{2}-1\right) \cdot 2 x}{1} \quad \text { [using L'Hospital Rule] } \\ &=2 \sec ^{2} 0=2 \end{aligned} $$

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

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