JEE Main 2018 (Online) 15th April Evening ShiftQuestion 1
Let $f(x)=\begin{cases}(x-1)^{\frac{1}{2-x}}, & x>1,\ x\neq2\\ k, & x=2\end{cases}$ The value of $k$ for which $f$ is continuous at $x=2$ is
Limits & Continuity
Solution
Type $1^\infty$: $\lim_{x\to2}\frac{x-2}{2-x}=-1$, so $f\to e^{-1}$ and $k=e^{-1}$.