Question

Evaluate: $\underset{x\to 1}{\lim }\left(\frac{\tan \left(x^2-1\right)}{x-1}\right)$

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

Answer 1

The correct option is A

$2$

Explanation for correct option:

Find the value of $\underset{x\to 1}{\lim }\left(\frac{\tan \left(x^2-1\right)}{x-1}\right)$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{\tan \left(x^2-1\right)}{x-1}\right) \\ I=\frac{\tan \left(1^2-1\right)}{1-1} \\ I=\frac{0}{0}\left[\text{indeterminent Form}\right] \end{gathered}$$

Using the L' Hospital rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{se{c}^2\left(x^2-1\right)2x}{1}\right) \\ \Rightarrow I=se{c}^2\left(1^2-1\right)2\times 1 \\ \Rightarrow I=se{c}^2\left(0\right)\times 2 \\ \Rightarrow I=1\times 2\left[\text{Where},se{c}^2\left(0\right)=1\right] \\ \therefore I=2 \end{gathered}$$

Hence, the correct answer is Option A.