Mathematics - Continuity - Differentiability Question with Solution | TestHub

Since $f(x)$ is continuous at $x=2$. $$\begin{array}{rl}& \therefore \underset{x\to 2}{\lim }f(x)=f(2)\\ & \Rightarrow \underset{x\to 2}{\lim }(x-1{)}^{\frac{1}{2-x}}=k\left(1^{\infty }\text{ form }\right)\\ & \therefore e^{l=k}\end{array}$$ where $$\begin{array}{rl}& l=\underset{x\to 2}{\lim }(x-1-1)\times \frac{1}{2-x}=\underset{x\to 2}{\lim }\frac{x-2}{2-x}\\ & =\underset{x\to 2}{\lim }\left(\frac{x-2}{x-2}\right)\\ & \Rightarrow k=e^{-1}\end{array}$$