Let $f\colon \mathbb{R} \to \mathbb{R}$ be the function
$$ f(x) = \begin{cases} \ x \exp\left(\frac{2x}{x^2-1}\right) &\text{if }x \in \mathbb{R} \setminus\{-1,1\},\\ \ 0 &\text{if }x = 1 \text{ or }x =-1. \end{cases}$$
Is this function continuous at $x=1$ and $x=-1$?
My guess:
I think that the values given to $f$ at $x=1$ and $x=-1$ are just hypothetical in case $f$ was continuous, but in reality $f$ is not continuous at these points, am I right?
But if what I just said is correct, what is the point of giving these values if f is actually not continuous at these points?