Question

Let $f\left(x\right)=\{\begin{array}{c}(x-1{)}^{\frac{1}{2-x}},x>1,x\ne 2\\ \\ k,x=2\end{array}$
The value of $k$ for which $f$ is continuous at $x=2$ is :

• A. $e^{-1}$
• B. $e^{-2}$
• C. $e$
• D. $1$

Answer 1

The correct option is A $e^{-1}$

$f\left(x\right)=\{\begin{array}{c}(x-1{)}^{\frac{1}{2-x}},x>1,x\ne 2\\ k,x=2\end{array}$
As $f\left(x\right)$ is continuous at $x=2$

$\therefore k=\underset{x\to 2}{lim}(x-1{)}^{\frac{1}{2-x}}$

$=\underset{x\to 2}{lim}{e}^{(x-1-1)\cdot \frac{1}{2-x}}$

$=\underset{x\to 2}{lim}{e}^{(-1)}$

$=e^{-1}$