If the function defined by $f\left(x\right)=\left\{\begin{array}{c}{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1}, \text{for }x\ne 2\\ k, \text{ for }x=2\end{array}\right.$ is right continuous at $x=2,$ then $k=$

Given:  the function defined by $f\left(x\right)=\left\{\begin{array}{c}{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1}, \text{for }x\ne 2\\ k,\text{ for }x=2\end{array}\right.$ is right continuous at $x=2$

So here $\mathrm{R}.\mathrm{H}.\mathrm{L}.=f\left(2\right)$

$\Rightarrow k=\underset{h\to 0}{\lim }f\left(2+h\right)$

$\Rightarrow k=\underset{h\to 0}{\lim }{\left({\left(2+h\right)}^2+e^{\frac{1}{2-\left(2+h\right)}}\right)}^{-1}$

$\Rightarrow k=\underset{h\to 0}{\lim }{\left({\left(2+h\right)}^2+e^{\frac{1}{-h}}\right)}^{-1}={\left(4+e^{-\infty }\right)}^{-1}=\frac{1}{4}\left(\because e^{-\infty }=0\right)$