0

I keep on seeing on the internet that $\lim_{x \to 0}(\sqrt{x}) = 0$. I'm skeptical, since $\sqrt{x}$ has no left-side limit, but I wasn't sure. So I did a proof involving the delta-epsilon definition of the limit that is taught in Calculus 1, which I am currently reviewing, and I got the result that $\lim_{x \to 0}(\sqrt{x}) \ne 0$. Here is my proof, informally:

In order to write that $\lim_{x \to 0}(\sqrt{x}) = 0$, we must show that for all $\epsilon > 0$, there exists $\delta > 0$ such that any x satisfying $0 < \lvert x \rvert < \delta$ also satisfies $\lvert \sqrt{x} \rvert < \epsilon$. Consider $x = -\delta/2$. $x = -\delta/2$ is an example of a value that satisfies $0 < \lvert x \rvert < \delta$, but does not satisfy $\lvert \sqrt{x} \rvert < \epsilon$. Hence, it is not necessarily true that any x satisfying $0 < \lvert x \rvert < \delta$ also satisfies $\lvert \sqrt{x} \rvert < \epsilon$. And we have shown this for any $\delta > 0$ and any $\epsilon > 0$. Therefore, for all $\epsilon > 0$, there is no $\delta > 0$ such that any x satisfying $0 < \lvert x \rvert < \delta$ also satisfies $\lvert \sqrt{x} \rvert < \epsilon$. This is in direct contradiction to the the delta-epsilon definition of a limit. Hence, we cannot write $\lim_{x \to 0}(\sqrt{x}) = 0$.

Where is my mistake?

Lauren S
  • 409

1 Answers1

3

The definition that was taught to you was, simply missing a part, which isn't emphasised as much in an elementary course, but becomes rather important when doing things in analysis. The main thing is, you don't worry about points outside of the domain of the function.

Here is a correct definition:

Let $f\colon A\to\mathbb R$ be a function, and $a$ be a limit point of $A$. If there is a real number $L$ such that for every $\varepsilon>0$, there exists $\delta>0$ such that $$x\in\left[(a-\delta,a+\delta)\setminus\{a\}\right]\cap A\implies|f(x)-L|<\varepsilon$$ then, we say that $\lim_{x\to a}f(x)$ exists and is equal to $L$.

(The $\cap A$ is important, as if $x\notin A$, then $f(x)$ does not make sense.)

In case it sounds technical to you, here is a simplified version, looking more like what you studied.

Let $f\colon I\to\mathbb R$ be a function, where $I$ is some interval and $a\in I$. If there is a real number $L$ such that for every $\varepsilon>0$, there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ for every $x\in I$ satisfying $0<|x-a|<\delta$, then, we say that $\lim_{x\to a}f(x)$ exists and is equal to $L$.

Again, the condition $x\in I$ is there just to make sure $f(x)$ makes sense.

In your case, $x\mapsto\sqrt x$ is a function from $[0,\infty)$ to $\mathbb R$. Then, the statement $\lim_{x\to0}\sqrt x=0$ becomes

For every $\varepsilon>0$, there exists $\delta>0$ such that $\sqrt(x)<\varepsilon$ for every $x\in[0,\infty)$ such that $0<|x|<\delta$. But that constraint on $x$ is the same as saying $0<x<\delta$, so $x=-\delta/2$ step is not valid.

Hope this helps. :)

ultralegend5385
  • 10,106
  • 1
    I would be careful with saying that OP made a mistake. Unfortunately, not every text is as rigorous as possible as it pertains to the definition of a limit (so that one-sided limits and more exotic situations can be explored). Some explicitly consider only points on the interior of intervals of definition. – Cameron L. Williams Apr 17 '24 at 16:13
  • Many beginning and calculus (as opposed to analysis) texts require the function to be defined on an open interval containing $a$, except perhaps at $a$, rather than the more general definition that is valid at any accumulation point of the domain, which is what you quote. – Arturo Magidin Apr 17 '24 at 16:21
  • @ArturoMagidin: Sure, but the problem stays the same. (Same for the comment by @CameronWilliams) – ultralegend5385 Apr 17 '24 at 17:17
  • 1
    @ultralegend5385: What doesn't "stay the same" is the assertion that the OP has "made a mistake". The OP may be faithfully following the definition that they are working with, in which case there is no "mistake". It is better to clarify that there are conflicting definitions, and that the answer depends on which definition is being used. So I would disagree with both the claim that there is a definite "mistake", and the claim that the problem "stays the same." – Arturo Magidin Apr 17 '24 at 17:55
  • @ArturoMagidin: Good point. My bad, I made some edits. – ultralegend5385 Apr 17 '24 at 18:34
  • @ultralegend5385 Ah thanks. I suspected that was what was wrong with the definition given to me. Only thing that still looks off about your definition is the requirement for $a$ to be within $I$. I was taught that $a$ doesn’t need to be within the function’s domain in order to take the limit. Suppose we were taking the limit as x approaches zero of a function that undefined at 0 but sqrt(x) anywhere else? Would the limit still be zero? – Lauren S Apr 17 '24 at 19:40
  • @LaurenS: Yes, that comes under the more technical definition I had written before. $0$ is a limit point of $(0,\infty)$. End points of intervals like $(a,b)$ are also limit points. – ultralegend5385 Apr 18 '24 at 03:58