Question

$\underset{x\to 2}{\lim }\left(\frac{{\sin }^{-1}\left(x-2\right)}{x^2-4}\right)=$

• A. $0$
• B. $2$
• C. $\frac{1}{4}$
• D. $1$

Answer 1

The correct option is C

$\frac{1}{4}$

Explanation for the correct option:

Find the value of $\underset{x\to 2}{\lim }\left(\frac{{\sin }^{-1}\left(x-2\right)}{x^2-4}\right)$:

The given equation Is,

$I=\underset{x\to 2}{\lim }\left(\frac{{\sin }^{-1}\left(x-2\right)}{x^2-4}\right)$

Rewrite the above Equation as

$$\begin{gathered} I=\underset{x\to 2}{\lim }\left(\frac{{\sin }^{-1}\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right) \\ \Rightarrow I=\underset{x\to 2}{\lim }\left(\frac{{\sin }^{-1}\left(x-2\right)}{\left(x-2\right)}\times \frac{1}{\left(x+2\right)}\right) \\ \Rightarrow I=1\times \frac{1}{\left(2+2\right)} \\ \Rightarrow I=\frac{1}{4} \end{gathered}$$

Hence, the correct answer is option (C).