Doubt on the range of a function while considering continuity

Question

Theorem: If $f$ is a continuous function on the closed interval $[a,b]$, then $f$ is uniformly continuous.

Now I try to apply this theorem on $f(x)=\frac{1}{(1-x)^2}$ in the interval $[0,2]$. It's not hard to show that the function is not uniformly continuous (intuitively near $x=1$ the function grows abnormally fast). Thus I concluded that $f(x)$ is not continuous in the interval $[0,2]$. But we can also observe that $\lim_{x \to 1^{+}} f(x)=+\infty=\lim_{x\to 1^{-}}f(x)$ and $f(1)=\infty$. Now I concluded that we do not allow infinity to be in the range of a function, otherwise this would be perfectly consistent with the definition of continuity. But this seems kind of contradictory to me since lots of time we say things like "$f(x)=+\infty$ at x= something". Do we say these things in the limit sense? I consulted some sites and books but none of them explicitly says that $f(c)$ has to take a finite value to be continuous at that point.

Answer 1

You are misinterpreting. If $f$ is a continuous function on a closed and bounded interval then it must be bounded. For your case, the function is unbounded at $x=1$ so the function cannot be continuous.

Answer 2

For continuity of a function at a point, first of all , limit of the function must exist at that point finitely and then it must be equal to the functional value.

So, Limit should exist at the point where a function is continuous.

And , without being continuity, we can't go for uniform continuity.