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

Answer 1

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

If $f\left(x\right)$ is continuous at $x=2$, then

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

Above is $1^{\infty }$ form,

$\therefore k=e^l$

where $l=\underset{x\to 2}{lim}(x-1-1)\times \frac{1}{2-x}=\underset{x\to 2}{lim}\frac{x-2}{2-x}=-1$

$⟹k=e^{-1}$