Is Sqrt X Continuous at 0

Alkatran · Dec 5, 2021

Is the square root of 0 undefined?

I was recently told that the square root of 0 is undefined because the limit of a square root didn't exist at 0. The reason is that from the negative direction you have i and from the positive direction you don't.

At first, i agreed with this. It made enough sense.

Then, about 5 seconds after i was out of the room, I pictured it on a complex graph (you know, y axis = imaginary, x axis = real) and noticed that the two points were getting closer together.

You could in fact say that .5i is closer to .5 than 1i is to 1. So, the limit does converge to 0 + 0i.

..right?

Hurkyl · Oct 24, 2004

This is all a red herring. Square roots are defined algebraically, not by limits. 0 is a solution to x^2 = 0, so 0 is a square root of 0.

[itex]\lim_{x \rightarrow 0} \sqrt{x}[/itex] does, in fact, not exist, but that's because [itex]\sqrt{x}[/itex] is undefined for x < 0. However, when you instead consider a complex square root function, the limit does exist and [itex]\sqrt{x}[/itex] is continuous at 0.

Alkatran

This is all a red herring. Square roots are defined algebraically, not by limits. 0 is a solution to x^2 = 0, so 0 is a square root of 0.
[itex]\lim_{x \rightarrow 0} \sqrt{x}[/itex] does, in fact, not exist, but that's because [itex]\sqrt{x}[/itex] is undefined for x < 0. However, when you instead consider a complex square root function, the limit does exist and [itex]\sqrt{x}[/itex] is continuous at 0.

So it's right and wrong depending on wether or not you consider the complex plane... the fact that "i" was used to disprove it makes it... true?

But I get it. Next time I'll just say it isn't defined and put a litte (in real number) next to it.

JasonRox · Nov 27, 2021

This is interesting. Does anyone else with ideas on the square root of 0?

Alkatran

Chalk one down to not reading "limit of" in the question. :blushing:

But it's good to know the question can go either way.

matt grime · Oct 25, 2004

No it isn't (interesting, I mean), or am I being dense?

sqrt is continuous as a function from R+ (the non-negative reals) to R, and from C to C, and even from R to C (all in the usual {metric} topology).

Since a "function" without a domain (and codomain) is not actually a function the question is easily answered.

Alkatran · May 1, 2017

No it isnt, or am I being dense?
sqrt is continuous as a function from R+ (the non-negative reals) to R, and from C to C, and even from R to C (all in the usual {metric} topology).

Since a "function" without a domain (and codomain) is not actually a function the question is easily answered.

Can you explain that a bit more, please?

http://www.vbforums.com/attachment.php?s=&postid=1821439 [Broken] : 3d graph with z axis as imaginary image, y as real image, and x as domain, the blue line is on z, and the red on y, they intersect, right?

I'll apologize for the ... roughness... of the graph. It's not an exact one, made with paint.

MiGUi

The square root of zero is zero.

I said to Mathematica to find the zeroes of [tex]\sqrt{x}[/tex] in the (-1,1) interval and the result was not exactly zero:

[tex] x \rightarrow -4.97297 \times 10^{-15} + 4.2265 \times 10^{-16} i[/tex]

Whats this?

Alkatran

Just to clarify, the question said "LIMIT OF" in it. So the question is: is the limit of the square root of x as x approaches 0 defined?

It isn't in the real numbers, but it is in the complex? Are (1,0) and (0,1) closer to each other than (2,0) and (0,2)?

matt grime

Just to clarify, the question said "LIMIT OF" in it. So the question is: is the limit of the square root of x as x approaches 0 defined?
It isn't in the real numbers, but it is in the complex? Are (1,0) and (0,1) closer to each other than (2,0) and (0,2)?

Yes it's defined, and yes it's zero.

It isn't a function from R to R, so if we're even going to talk about having square roots of negative numbers, then we must mean from R to C, where it is obviously continuous at 0 since for any sequence of real numbers tending to zero, the square root tends to zero (in the metric and any other reasonable topology).

If you only want real output then the input can only be positive, and thus one cannot speak of taking a limit with a sequence of negatives.

matt grime

Here is an interesting one though: consider the square root of e^{i\theta} as theta increases from 0 to 2pi. As you go round you find the square root of 1 from below is -1... how do we get round that?

Alkatran

Here is an interesting one though: consider the square root of e^{i\theta} as theta increases from 0 to 2pi. As you go round you find the square root of 1 from below is -1... how do we get round that?

That's right, the graph should have 2 red and 2 blue lines. They still all intersect at 0, though.

jcsd

I suppose limits don't really tell us anyhting about whether or not a function is defined at a point for example the function f(x) = 1 if x is irrational, f(x) = 0 if x is rational is defined for all x in R yet it is discontinous about every point! Yet the limit of f(x)= x/x exists for x = 0, but the function is not defined at x = 0 (and hence there is a removable discontinuity at zero).

The interesting thing about limits in the complex numbers is that whilst in the real numbers for a limit to exist we say that the limit must be the same from both sides, in the complex numbers for a limit to exist it must be the same from all directions (and there are an infinite amount of directions!).

jcsd

Whilst we're omn the subject, if we're taking the limit of a function whose domain closed subset of the reals (such as the non-negative numbers), do we still require that for the limit to exist at a point that both the left and right-sided limit exists?

matt grime · Oct 25, 2004

Only if sequences of numbers approaching from the left and right are all members of the domain.

Some phenomena:

the function from R to R that is 1 for non-negative x, and zero otherwise is continuous when considered as a function with domain restricted to the no-negatives. (compare and contrast to the idea of a subspace topology), it is after all then just the function f(x)=1 for all x in the domain, which had better be continuous.

also note that something like f:Q \to Q, f(x)=1/(x^2-2), is continuous.

though surely it is better, if possible, to think of continuity in terms of inverse images of open sets.

Is Sqrt X Continuous at 0

Square root of 0 undefined?

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