How to make the function $f(x)= x^{{(x-2)}^{-1}}$ continuous at $x=2$

Question

The function $f(x)= x^{{(x-2)}^{-1}}$ is defined for all $x>0$ and $x \neq 2$ How should $f(2)$ be defined so that $f$ is continuous on $(0,\infty)$

By graphing the function I can see that it is discontinuous at $x=2$

I would imagine that $f(2)$ should be defined as $0$, how can I explain that reasoning mathematically?

Answer 1

You cannot make your function continuous at $x=2$, the discontinuity is not removable. This is seen from finding LHL and RHL at $x=2$. LHL is simply $0$ because we get limit of the form $2^{-\infty}$ and RHL is $\infty$ as we get $2^{\infty}$ form.