# Does $\sqrt{x}$ have a limit for $x \to 0$?

I am taking a calculus course, and in one of the exercises in the book, I am asked to find the limits for both sides of $\sqrt{x}$ where $x \to 0$.

Graph for sqrt(x) from WolframAlpha:

This is how I solved the exercise:

For simplicity, I choose to disregard the negative result of $\pm\sqrt{x}$. Since we are looking at limits for $x \to 0$, both results will converge at the same point, and will thus have the same limits.

$\sqrt{x}$ = $0$ for $x = 0$.

$\sqrt{x}$ is a positive real number for all $x > 0$.
$\displaystyle \lim_{x \to 0^+} \sqrt{x} = \sqrt{+0} = 0$

$\sqrt{x}$ is a complex number for all $x < 0$.
$\displaystyle \lim_{x \to 0^-} \sqrt{x} = \sqrt{-0} = 0 \times \sqrt{-1} = 0i = 0$

The solution in the book, however, does not agree that there exists a limit for $x \to 0-$.

I guess there are three questions in this post, although some of them probably overlaps:

• Does $\sqrt{x}$ have a limit for $x \to 0$?
• Are square root functions defined to have a range of only real numbers, unless specified otherwise?
• Is $\sqrt{x}$ continuous for $-\infty < x < \infty$?

WolframAlpha says the limit for x=0 is 0: limit (x to 0) sqrt(x)

And also that both the positive and negative limits are 0: limit (x to 0-) sqrt(x)

If my logic is flawed, please correct me.

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By the way, "a positive complex number" does not make sense: there is no ordering on the field of complex numbers, and so it does not make sense to talk about positive or negative complex numbers. –  M Turgeon Sep 10 '12 at 0:52
I added \displaystyle to the places where you used limits. This forces the typesetting to place the x\to 0 below the limit notation. –  wckronholm Sep 10 '12 at 1:08
Thanks, that looks a lot better. I also removed the "positive complex number" remark. –  CheeseSucker Sep 10 '12 at 1:17

The answers to your questions depend on whether you are working with $\mathbb{R_{\geq 0}}$ or $\mathbb{R}$ as your domain. $\sqrt{x}$ is really not the same function in these two cases.

With a domain of $\mathbb{R_{\geq 0}}$:

• Clearly $\lim_{x \to 0^+} \sqrt{x} = 0$. However $\lim_{x \to 0^-} \sqrt{x}$ is not defined for negative numbers in this case, so $\lim_{x \to 0^-} \sqrt{x}$ is undefined.
• If your course is on real analysis, they most likely assume that the domain is $\mathbb{R_{\geq 0}}$. However, this could depend very much on the context. If you are unsure what your course's convention is, you should consult a teaching assistant.
• In this case, $\sqrt{x}$ is not continuous on $\mathbb{R}$, since it is not even defined everywhere.

With a domain of $\mathbb{R}$:

• As your plot shows you, both the real and the imaginary parts tend to 0, and so $\sqrt{x}$ tends to 0.
• See above.
• Once again, your plot shows you that both the real and imaginary parts are continuous functions over $\mathbb{R}$, and so $\sqrt{x}$ is continuous over $\mathbb{R}$.

Note that there is an even more general case, where the domain is $\mathbb{C}$. This is where things get very strange.

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I just want to point out that you shouldn't trust graphs blindly. Compare: tinyurl.com/8hbncsm and tinyurl.com/8lk9nog. Both are of the same function, but with different zoom levels. The function is not defined for x = 0, and the artifacts you see are due to precision / rounding errors. –  CheeseSucker Sep 10 '12 at 1:30

There is undeniably a right-hand limit. You have $\sqrt{x}\to 0$ as $x\downarrow 0$. In fact, since $\sqrt{0} = 0$, you know the square root function is right continuous at zero.

However, there is no possibility limit as $x\to 0-$, since the domain of this function is $[0,\infty)$. To discuss the left-hand limit of a function at a point $a$, the function must be defined in some interval $(q, a)$, where $q < a$.

The square root function is continuous on its domain. Since it is not defined on $(-\infty, 0)$, it is often informally said that it has a "limit at zero."

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Is the domain of square root functions always $[0, \infty)$ unless explisitly noted otherwise? After all, $i$ is defined as $\sqrt(-1)$. So I would think that, by default, the domain of a function would be whatever values it is defined for? –  CheeseSucker Sep 10 '12 at 0:32
If you are in the real domain, as you would be in a Calculus class, yes. The square-root function in the complex domain is a little more complicated. There is a square root function on any domain in the complex plane that is simply connected (every loop can be shrunk to a point in the domain). Beware, there are weirdisms here. –  ncmathsadist Sep 10 '12 at 0:39