How to apply the continuity definition to $f(x)=\sqrt{x}$?

Question

According to the continuous definition (at least the one I know)

A function is continuous if it is continuous at every point of its domain.

The problem I have is that in order to have continuity at a point both one sided limits must exist and have the same value as the function evaluated at that particular point, but what about functions whose domain has an endpoint? Are all of these functions discontinuous because of the lack of a one sided limit or am I missing something?

My particular problem with this is to decide wether the function $f(x)=\sqrt{x}$ is continuous at $x=0$ or not because of the lack of a left hand limit at $x=0$.

See this question: Is $\sqrt x$ continuous at $0$? Because it is not defined to the left of $0$ — Scene, Jun 23 '23 at 02:20
Does this answer your question? Is $\sqrt x$ continuous at $0$? Because it is not defined to the left of $0$ — John Omielan, Jun 23 '23 at 02:22
The continuity definition requires the intersection of the domain of $f$ with a neighborhood of $0$. Hence, for $\sqrt{x}$ the limit at $0$ coincides with the limit at $0^+$. — Bernkastel, Jun 23 '23 at 02:52
We only talk about continuity on a domain where the function exists. The function doesn't exist if $x$ is negative, so it doesn't make sense to talk about limits for negative values of $x$. — Accelerator, Jun 23 '23 at 07:21

score 0 · Answer 1 · answered Jun 23 '23 at 03:01

By definition, a function is continuous at a point $x$ if and only if $\lim_{c\to x} f(c)=f(x)$. Clearly, as $x$ approaches $0$ (from the right side), it gets smaller and smaller and so $\sqrt{x}$ approaches $0$. Since $\lim_{x\to0} \sqrt{x}=0=\sqrt{0}$, so $\sqrt{x}$ is continuous at $x=0$.

How to apply the continuity definition to $f(x)=\sqrt{x}$?

1 Answers1