Question

$\displaystyle\lim_{x \to 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ is equal to

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

Answer 1

Answer: (B)

We have, $\displaystyle\lim_{x\to1} \frac{\tan\left(x^{2}-1\right)}{x-1} $ $\left(\frac{0}{0} \text { form }\right) $
$= \displaystyle\lim_{x\to1} \frac{\sec^{2} \left(x^{2} -1\right) .2x}{1}$
[using L’ Hospital’s rule]
$ = 2. \sec^{2} \left(0\right) = 2 $