Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}(x-1{)}^{\frac{1}{2-x}},& x>1,x\ne 2\\ k,& x=2\end{array}$. The value of k for which f is continuous at x = 2 is :-

To ensure the function $f(x)=(x-1{)}^{\frac{1}{2-x}}$ is continuous at $x=2$, we must define the value $k$ such that the limit of the function as $x$ approaches $2$ equals the function value at that point.

Mathematically, we require:

This condition serves as our guiding star for resolving the discontinuity.

Before we can determine $k$, we must evaluate the limit as $x$ approaches $2$. Substituting $x=2$ directly into the expression, the base $(x-1)$ approaches $1$, while the exponent $\frac{1}{2-x}$ approaches $\infty$.

This results in the classic indeterminate form $1^{\infty }$. In the JEE arena, recognizing this form is the critical first step toward the solution.

To resolve this indeterminate form, we utilize the standard limit formula for $1^{\infty }$ expressions. For a limit of the form ${\lim }_{x\to a}[g(x){]}^{h(x)}$, where $g(x)\to 1$ and $h(x)\to \infty$, the limit is equivalent to:

In our specific case, we identify $g(x)=x-1$ and $h(x)=\frac{1}{2-x}$. Substituting these into our formula, we obtain:

We now simplify the expression within the exponent. The term $(x-1-1)$ simplifies to $(x-2)$.

The limit expression becomes:

Since $(2-x)=-(x-2)$, we can cancel the $(x-2)$ terms, provided $x eq 2$. This yields:

Consequently, the entire limit evaluates to $e^{-1}$. Therefore, the value required to fill the hole and restore continuity is $k=e^{-1}$ (or $\frac{1}{e}$).