Is the function
$$f(x)=\frac{x-2}{x(x-2)}$$
continuous?
I'm in high school and calculus is something that is not yet very comfortable to me. I read that we call this function continuous in its domain.
What does it exactly mean, because…
The points $0, 2$ are clearly not in its domain, so the definition of continuous function seems a bit weird $\lim_{x \to a^+}f(x) = \lim_{x \to a^-}f(x) = f(a) $ but for points $0,2,$ $\;f(a)$ is not valid as they are not in the domain, so how are we checking the continuity there. Also by this definition, any function of form $\frac{anything}{0}$ is continuous then…
Also it seems a bit counter intuitive as at $0,\;$ $\lim_{x \to 0^+}f(x) =\infty, \lim_{x \to 0^-}f(x) = -\infty $ and yet we are calling it continouos.
Also what exactly is the difference between a "function being continouos" and a "function being continouos in its domain".
Best Answer
You are correct. The function is continuous, but the idea of continuity only applies at points in the domain. The points $0$ and $2$ are not in the domain.
You are also right to be "worried" about those points.
The function can be extended to a function $\hat{f}$ including $2$ in its domain (you must decide how to define $\hat{f}(2)$ in that case; the only value that results in a function that is still continuous is to define $\hat{f}(2)=\tfrac12$).
You can also extend this to an even larger extension $f^*$ which includes $0$ in its domain, but it will not be continuous no matter how you define $f^*(0)$ for the reasons you point out.
Part of what you should take away from this is that the domain needs to be explicit before you can say whether a function is continuous. A function can be discontinuous, but restricting its domain can result in a continuous function. And a function can be continuous, but it can have continuous or discontinuous extensions. The extensions/restrictions really are different functions.