Are root functions continuous given any domain?

Question

For instance take $f(x)=\sqrt{4-x}$. For all cases, we can say that the function is continuous at $x=a$ if $\lim\limits_{x\to a}f(x)=f(a)$. However, if a limit exists it is unique. Thus, the limit of $f(x)$ does not exist for any $x=a$ because $\sqrt{a}=\pm b$ (where $b$ is a positive real number), and thus the graph function is discontinuous. From what I have seen people tend to disregard the $-b$. In that case, the $\lim\limits_{x\to a}f(x)=f(a)$ for any $a>0$. Thus the graph would be continuous everywhere (given the proper domain) but at $x=0$.

I'm just a little confused. So if someone could explain how this works I would greatly appreciate it.

Answer 1

The radical sign represents the principal or nonnegative square root. The $-b$ isn't being ignored: it's just negative (if $b$ is a positive real number, as you say) and thus it isn't $\sqrt{a}$ but is instead $-\sqrt{a}$.

As a result, when you have the function $f(x)=\sqrt{4-x}$, your outputs are going to be nonnegative because the radical sign represents the principal square root of $4-x$.

Answer 2

We can check if every $n$th root function is continuous in a slightly sneaky way by making use of a version of the inverse function theorem, which states the following:

If $I\subseteq\mathbb{R}$ is an interval and $f:I\to\mathbb{R}$ is injective and continuous on $I$, then the function $f^{-1}:I\to f(I)$ is continuous.

Consider the function $f:[0,\infty)\to [0,\infty)$ defined as $f(x)=x^n$ for some $n\in\mathbb{N}$. It is not too difficult to show that all polynomial functions are continuous. It is also good to note that $f$ is bijective on the specified domain and codomain, so $f$ satisfies the hypothesis of the version of the inverse function theorem stated above. This implies that $f^{-1}:[0,\infty)\to[0,\infty)$ is continuous, that is $f^{-1}(x)=\sqrt[n]{x}$ is continuous on $[0,\infty)$.