Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:
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.

share|improve this question
    
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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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."

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.