If a one-sided limit of a function doesn't exist, how can a function

[Eclair_de_XII]

Homework Statement

"Let $f(x)=\sqrt{x}$. Prove or disprove $\lim_{x\rightarrow0} \sqrt{x}=0$."

Relevant Equations

A function $f$ has a limit $L$ at $a$ iff $\lim_{x\rightarrow a^-}=L=\lim_{x\rightarrow a^+}$.

Instinct tells me to just plug in the number, say the limit is zero, and be done with it. But at the same time, while reading the statement from the "Relevant equations" section of this post, I cannot feel but feel some doubt as to whether or not this is the right approach. I mean, only the right-sided limit is defined, while the left-sided limit isn't. So am I to conclude that since the root function doesn't even have a left-handed limit, that I don't need to worry about the statement in the "Relevant equations" section?

Sorry; it's been many years since I last took Calculus I, so I am sorry for being dense when it comes to simple concepts like this.

 

[Mark44]

Problem Statement: "Let $f(x)=\sqrt{x}$. Prove or disprove $\lim_{x\rightarrow0} \sqrt{x}=0$."
Relevant Equations: A function $f$ has a limit $L$ at $a$ iff $\lim_{x\rightarrow a^-}=L=\lim_{x\rightarrow a^+}$.

Instinct tells me to just plug in the number, say the limit is zero, and be done with it. But at the same time, while reading the statement from the "Relevant equations" section of this post, I cannot feel but feel some doubt as to whether or not this is the right approach.

It isn't. What's usually meant when you're asked to prove a limit is to do so using a $\delta - \epsilon$ argument. And "plugging in a number" works only for the simplest of functions. For example, $\lim_{x \to 1}\frac {x^2 - 1}{x - 1}$ exists, but you can't just "plug in 1" for x.

I mean, only the right-sided limit is defined, while the left-sided limit isn't. So am I to conclude that since the root function doesn't even have a left-handed limit, that I don't need to worry about the statement in the "Relevant equations" section?

On the contrary, your relevant definition is very important here. If the right-side limit exists, but the left-side limit doesn't exist, what does that say about the existence of the two-sided limit?

Sorry; it's been many years since I last took Calculus I, so I am sorry for being dense when it comes to simple concepts like this.

 

[Eclair_de_XII]

what does that say about the existence of the two-sided limit?

So the two-sided limit does not exist? But doesn't that contradict the continuity of $\sqrt{x}$ at $x=0$?

 

[Orodruin]

Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.

 

[Eclair_de_XII]

it is therefore not continuous on $\mathbb{R}$.

Oh! If I restrict the domain of $f$ to non-negative real numbers, will the condition stated in the "Relative equations" section then be a non-issue?

 

[Orodruin]

Oh! If I restrict the domain of $f$ to non-negative real numbers, will the condition stated in the "Relative equations" section then be a non-issue?

If you also restrict the codomain.

The concept of continuity on $\mathbb R$ is a special case of the general topologigal definition.

 

[Mark44]

Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values

I assumed that the OP was talking about the real-valued square root function.

 

[Eclair_de_XII]

If you also restrict the codomain.

Okay, so it's continuous if I restrict the codomain and domain to the non-negative real numbers. Wait, but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?

 

[MidgetDwarf]

So the two-sided limit does not exist? But doesn't that contradict the continuity of $\sqrt{x}$ at $x=0$?

continuity and limit of a function are not equivalent. Recall that the definition of limits of a function requires that the number x is approaching be a limit point of the domain of the function. Just because a number is a limit point of a set, it does not imply that the limit point be in the set (think of open sets for the trivial case).

On the other hand, the definition of continuity states to consider if a function is continuous at a point, the minimum is that the point must be the domain of the function.

 

[MidgetDwarf]

Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.

However, it is defined if we restrict the domain to all real number greater than or equal to x. It is very trivial to show this.

 

[MidgetDwarf]

I think you should review the concept of limit points, open, closed, compact set, limit of a function, definition of continuity.

 

[WWGD]

...but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?

No, notice $f(x)=f(-x)$

 

[Mark44]

... but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?

No, notice $f(x)=f(-x)$

How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?

 

[WWGD]

How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?

Instead of considering both the positive and negative root, just ignore the negative root, that's all.

 

[PeroK]

Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.

This is very much the physicist's view. In real analysis, however, the mathematician cannot extend the concept of a number to the complex field. A physicist is, of course, free to adduce the complex numbers in any case.

Instead, the key point missed by the OP is that the formal definition of continuity restricts any points under consideration to the domain of the function.

In this case the "left-sided" limit is not valid at the point $x =0$. And, in fact, the limit and the right-sided limit are equivalent at this point.

 

[Mark44]

How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?

Instead of considering both the positive and negative root, just ignore the negative root, that's all.

This still makes no sense. The real-valued square root function doesn't have negative roots.

 

[WWGD]

This still makes no sense. The real-valued square root function doesn't have negative roots.

Yes, I phrased it poorly , yet many people consider the possibility of two roots notwithstanding the fact that an expression with two outputs cannot be a function. I thought this is what poster was referencing.

 

[Mark44]

yet many people consider the possibility of two roots

We've had many threads here at PF about this misconception.

 

[WWGD]

We've had many threads here at PF about this misconception.

Maybe we can match the misconception with Mr conception, Dave or otherwise ( I have a Venezuela connection so I know of some of the famous players) :).

 

[member 587159]

On the contrary, your relevant definition is very important here. If the right-side limit exists, but the left-side limit doesn't exist, what does that say about the existence of the two-sided limit?

I'm sorry but this response is as wrong as it can get.

The only definition of limit is the $\epsilon-\delta$ definition or something equivalent with it. The "definition" the OP is using is a theorem that holds under certain conditions. Here, these conditions are not satisfied. The notion of left limit simply doesn't make sense for the function $[0,\infty[\to \mathbb{R}: x\mapsto \sqrt{x}$.

If you would apply the $\epsilon-\delta$ definition of limit to this function, you can show that the limit exists and equals $0$.

It is certainly wrong to say the limit does not exist, as you seem to be suggesting.

 

[Mark44]

I'm sorry but this response is as wrong as it can get.

The only definition of limit is the $\epsilon-\delta$ definition or something equivalent with it. The "definition" the OP is using is a theorem that holds under certain conditions. Here, these conditions are not satisfied. The notion of left limit simply doesn't make sense for the function $[0,\infty[\to \mathbb{R}: x\mapsto \sqrt{x}$.

If you would apply the $\epsilon-\delta$ definition of limit to this function, you can show that the limit exists and equals $0$.

It is certainly wrong to say the limit does not exist, as you seem to be suggesting.

What I'm saying is that the two-sided limit, $\lim_{x \to 0} \sqrt x$, doesn't exist. Clearly $\lim_{x \to 0^-} \sqrt x$ does not exist, so the two-sided limit can't exist.

 

[member 587159]

What I'm saying is that the two-sided limit, $\lim_{x \to 0} \sqrt x$, doesn't exist. Clearly $\lim_{x \to 0^-} \sqrt x$ does not exist, so the two-sided limit can't exist.

Define two-sided limit please.

The limit $\lim_{x\to 0}\sqrt{x}$ does exist.

With this, I mean that for every $\epsilon>0$, there is a $\delta >0$ such that if $x\geq 0$ satisfies $0 <|x| <\delta$, then $|\sqrt{x}|<\epsilon$.

 

[Mark44]

Define two-sided limit please.

I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.

 

[PeroK]

I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.

Yes, but that doesn't apply to an end point on a closed interval. The left sided limit if you allow it to be valid is vacuously any number you want. There are simply no qualifying values of $x <0$ in the domain of the function.

 

[member 587159]

I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.

I think you will be interested in reading this. It explains why we can't apply this theorem here. We need the point at which the limit is taken to be an interior point of the domain.

https://math.stackexchange.com/q/2637061/661543

 

[Orodruin]

If you also restrict the codomain.

So I was, of course, confusing myself here. You do not need to restrict the codomain. The topological definition of continuity is that the preimage of any open set in the codomain is open in the domain. In this particular case, the base elements of the standard topology on $\mathbb R^+$ that contain $x = 0$ are of the form $[0,a)$ where $a\in \mathbb R^+$. For the standard topology on the codomain $\mathbb R$, base elements that have $x = 0$ in their preimage are of the form $(-a,b)$, where $a, b > 0$ and the preimage of such a set is $[0,b^2)$, which is open in $\mathbb R^+$ and the function is therefore continuous at $x = 0$.

Edit: Also note that the $\epsilon$-$\delta$ definition of continuity is equivalent to the topological definition of continuity in metric topologies, such as the standard topology on $\mathbb R$. Of course, if you put some other topology on $\mathbb R$, you might have different results in general although it is worth pointing out that $\sqrt{x}$ in particular is continuous at $x = 0$ in both the trivial topology (the preimage of $\emptyset$ is $\emptyset$ and the preimage of $\mathbb R$ is $\mathbb R^+$) as well as in the topology where all sets are open (this is obvious, if all sets are open then all of the preimages are open by definition).