Let $g(x)=\frac{x^2-4}{x-4}$ Is $g(x)$ continuous at $x_0=4$?

Question

Let $g(x)=\frac{x^2-4}{x-4}$ Is $g(x)$ continous at $x_0=4$?

So first of all i wanted to ask. Why does this question make sense at all ? $x_0=4$ is not in the domain of $g(x)$ so asking about continuity makes no sense to me at this point?

Anyways $\lim_{x\to x_0}g(x)$ goes to infinity. So i assume its not continuous. Can i follow this using the epsilon delta definition of continuity ? Or is anything that goes to infinity directly not continuous ?

Answer 1

No, it is not since it is not defined at $4$.

Answer 2

g(x) is not defined at x=4 as the denominator is 0, therefore it cannot be continuous.

Answer 3

I want to elaborate on this a little bit. In most textbooks, they often have the function: $f(x) = \dfrac{x^2-16}{x-4}$ and $x = 4$ is the removable discontinuity as you know since the limit does not exist there. You can make the function continuous at $x = 4$ by define $f(4) = \displaystyle \lim_{x \to 4} f(x) = 8$. I hope this explanation gives you more insight into the question you are having.