Q84 / 180 · KCET 2007 · Mathematics

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

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

Solution

We have, $\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1} \quad\left(\frac{0}{0}\right.$ form $)$ $=\lim _{x \rightarrow 1} \frac{\sec ^{2}\left(x^{2}-1\right) \cdot 2 x}{1}$ [using L'Hospital's rule] $=2 \cdot \sec ^{2}(0)=2$

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

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