This year, in one of the exercises for university admission in the subject of Mathematics in Spain, it states the following:
Given the real function of a real variable defined over its domain as
$$f(x) = \begin{cases} \frac{x^2}{2+x^2} & \text{if} & x \leq -1 \\ \frac{2x^2}{3-3x} & \text{if} & x > -1 \end{cases}.$$
And I'm asked to study the continuity of the function in $\mathbb{R}$. My question is, if we realize that the domain of the function is $Dom(f) = \mathbb{R} - \{ 1 \}$, how should I answer the question that it's not continuous at $x=1$? This point doesn't belong to the domain, so it doesn't make sense to study continuity there, right? Some professors calculate the one-sided limits to realize that there's a vertical asymptote at $x=1$. What would be the formal definition of continuity for a function, from a topological perspective?
Any thoughts or suggestions you'd like to share would be greatly appreciated. Thank you very much.
You are right, the domain of $f$ is $\mathbb R \setminus \{1\}$. Thus it does not make sense to ask whether it is continuous at $x = 1$. But it does make sense to ask whether it is continuous at $x = -1$. Certainly the left-sided limit exists and and is $\lim_{x \to -1^-}f(x) = 1/3 = f(-1)$. Similarly the right-sided limit exists and is $\lim_{x \to -1^+}f(x) = 2/6 = 1/3 = f(-1)$. Thus $f$ is continuous at $x=-1$. Continuity in all other point of the domain is obvious.