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Prove that $f(x)=\sqrt{x}$ is continuous at $x=2$

Proof:

Givens: $\varepsilon > 0$ and $\delta > 0$ and $|x-2| < \delta$ for $\delta < 1$. Let $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ $$ |f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1 $$ $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$ Also, using the givens, $$ |x-2| < \delta < 1 \implies |x-2| < 1 \implies 1 < x < 3 \tag3 $$ Then it follows that $$ x - 2 < |\sqrt{3} + \sqrt{2}|\varepsilon \tag4 $$ Therefore, $$ |f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}} < \varepsilon. \tag5 $$ The reason I am unsure is because I know that delta should be minimized but I am unsure if all my steps are correct.

miracle173
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thewhale
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  • You shall know the difference between limit existing and continuous. – MathArt Mar 10 '24 at 20:57
  • How did you know that $$\frac{\delta}{\sqrt{3}+\sqrt{2}}< \epsilon$$ I can take e.g $\delta=\frac{1}{2}$ and $\epsilon= 10^{-1000}$ – Marco Mar 10 '24 at 20:58
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    Given an $\epsilon > 0$, you need to choose a $\delta$ such that whenever $|x - 2| \leq \delta$, $|f(x) - f(2)| \leq \epsilon$. – sudeep5221 Mar 10 '24 at 21:01
  • You go wrong at the first step: "Givens: $\epsilon > 0$ and $\delta > 0$ and..." Only $\epsilon$ and $x$ are given. – TonyK Mar 10 '24 at 21:13
  • is this the only wrong part? what else can be improved? – thewhale Mar 10 '24 at 21:14
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    Honestly, it seems like you are parroting a bunch of steps without comprehending why and what they mean. All your steps are there and there are no miscalculations. But the presentation is pure word salad and, as an argument, is gooble-dee-gook. I think as an excercise you need to be able to give an argument to someone who has never heard of a delta-epsilon proof and doesn't know the definition of continuous and explain it in a way they would understand. And understand why. – fleablood Mar 10 '24 at 23:01
  • how can this be achieved? I am not sure because I am very new to proofs. Actually I did not copy any steps but i did not put any effort into writing the words either I just assumed I only need to write the calculation. – thewhale Mar 10 '24 at 23:13
  • @thewhale Do you have a good understanding of the concept of continuity via its epsilon-delta definition? – Gary Mar 10 '24 at 23:49
  • @Gary I think so because I did this from the basis of Apostol calculus volume 1. he uses neighborhood and limits to explain it. perhaps I dont fully understand how this is connected to limits. I am not sure where I could be missing information. – thewhale Mar 10 '24 at 23:56

3 Answers3

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A general way to choose $\delta$ for any invertible function is the following: You want to show that for each $\epsilon>0$ there is $\delta>0$ such that, for each $x \in (2-\delta, 2+\delta)$ $$ \sqrt{2}-\epsilon<\sqrt{x}<\sqrt{2}+\epsilon $$ If $\epsilon>\sqrt{2}$ you can verify that the inequality is satisfied for any $\delta < (\epsilon+\sqrt{2})^2-2$. So we can assume $\epsilon \le \sqrt{2}$. By squaring all the members (that are all non-negative) $$ (\sqrt{2}-\epsilon)^2<x<(\sqrt{2}+\epsilon)^2 $$ So you want $$ 2-\delta >(\sqrt{2}-\epsilon)^2 $$ $$ 2+\delta < (\sqrt{2}+\epsilon)^2 $$ i.e. $$ \delta < \epsilon(2\sqrt{2} -\epsilon) $$

Marco
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    Careful: $a<b$ does not implies that $a^2<b^2$ – azif00 Mar 10 '24 at 21:13
  • @azif00 All the quantities here are positive, so it does, but i should have specified it – Marco Mar 10 '24 at 21:14
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    $\sqrt{2}-\epsilon$ is not necessarily positive. – azif00 Mar 10 '24 at 21:15
  • please explain how can I implement this into a proof – thewhale Mar 10 '24 at 22:24
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    TBH, while OP's proof had some issues (esp. before the edits), the overall strategy seems more straightforward and less prone to running into corner cases than this one. – Erick Wong Mar 10 '24 at 22:34
  • @ErickWong but this strategy actually works for any invertible function. I don't see an easy way to generalize the OP strategy any further. – Marco Mar 10 '24 at 22:38
  • One error-prone corner is that for large $\epsilon > 2\sqrt2$, the answer doesn't give any positive $\delta$. – peterwhy Mar 10 '24 at 22:41
  • thewhale: what do you mean? This is actually a proof that $$\lim_{ x \to 2} \sqrt{x}=\sqrt{2}$$ which part do you think can't be implemented in a proof? $$$$ @peterwhy The case in which $\epsilon>\sqrt{2}$ is treated separately, but usually one can restrict theirself to small $\epsilon$, as limits are local objects – Marco Mar 10 '24 at 22:45
  • @Marco no i just dont understand this in a way that i can use it to fix my original post – thewhale Mar 10 '24 at 23:15
  • @thewhale This is an independent argument. It can be used to find $\delta$ when you caculate the limit of an invertible function (as it is $\sqrt{x}$), but the $\delta$ that you have actually found in your post works as well. Your proof, after the edits is correct. And it actually proves that $\sqrt{x}$ is continuous in $x=2$. – Marco Mar 10 '24 at 23:39
  • @Marco the OP posts a „proof“ and asks if his reasoning is acceptable. You ignore this question and post your own proof. That does not make sense to me. So a downvote from me. – miracle173 Mar 11 '24 at 04:05
  • @miracle173 the main problem the OP had was in the choose of the "right" $\delta$ in the $\epsilon-\delta$ proof. The OP corrected it soon after. – Marco Mar 11 '24 at 07:45
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    @Marco No, the main problem that the OP has is that the proof is wrong. $|f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}} $ does not hold. – miracle173 Mar 12 '24 at 08:37
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For me you proof is very confusing and it is wrong. It is wrong because the equation $|f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}}$ does not hold. $|f(2.1)-f(2)|\approx 0.034$, but $\frac{0.1}{\sqrt3+\sqrt2}\approx 0.032$.

But now to the confusion. You state

Givens: $\varepsilon > 0$ and $\delta > 0$ and $|x-2| < \delta$ for $\delta < 1$.

and then you state

Let $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$

But what is the difference? The inequalities $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ is also given. And if $\delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ then $\delta < 1$ is redundant. So you can leave out this statement. Finally you do not have to prove that there is an interval of $\delta$ where your statment holds, it is sufficient that for each $\varepsilon$ you find on $\delta$. From you inequalities I deduce that $\min \{\frac12, \varepsilon\}$ is such a $\delta$, so I would write

Given $\varepsilon>0$ choose $\delta=\min{\frac12, \varepsilon}$ and assume that $|x-2|<\delta$.

Then you state:

$$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$ $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$

And what's that? I think you want to tell us

We have to show: $$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$

I think you want to tell us that and not that this is a prerequisite. So I think you should state this explicitly.

An what is $(2)$? We have

$$(\sqrt x-\sqrt 2)(\sqrt x+\sqrt 2)=(x-2)$$

This is formula is crucial to the proof so I would explicitly mention it explicitly.

From this and $(1)$ follows $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$

$(3)$ and $(4)$ follow immediately from the definition of $\delta$ and the $|x-2|<\delta$. I do not understand why you write "Then it follows" to justify $(4)$. Because it does not follow from $3$ Finally you do not tell us how you get inequality $(5)$. As I mentioned at the beginning. This inequality is wrong.

Here is a short proof:

Assume that $\varepsilon>0$ and $\delta=\varepsilon.$ If $|x-2|<\delta$ then $$|f(x)-f(2)|=|\sqrt x-\sqrt 2|=\frac{|\sqrt x-\sqrt 2|\cdot|\sqrt x+\sqrt 2|}{|\sqrt x+\sqrt 2|}=\frac{|x-2|}{\sqrt x+\sqrt 2}\lt\frac{\delta}{\sqrt 2}<\varepsilon$$ So $f(x)=\sqrt x$ is continuous at $x=2.$

miracle173
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  • im not sure what you mean. it clearly says that delta is min which depends on epsilon. you probably misunderstood that I was not looking to write out the entire proof with all the specific texts and exact steps I did on paper. I just needed to see if the reasoning is correct. it seems to follow closely with this post: https://math.stackexchange.com/questions/560307/prove-that-sqrtx-is-continuous-on-its-domain-0-infty. maybe some more clarification on your part is necessary – thewhale Mar 22 '24 at 02:30
  • I think your proof is wrong as I showed in the first paragraph of my anwer. Did you already check this? – miracle173 Mar 25 '24 at 14:37
  • if this is the case, suggest what would make it correct. the linked post is actually kind of similar to this one so I am also interested to see why it is correct and this one is not – thewhale Mar 25 '24 at 15:29
  • @thewhale I added a short proof. But why do you write ‚if this is the case…‘? Can‘t you see that this is the case? – miracle173 Mar 25 '24 at 17:04
  • Because im referencing the linked post. However one thing I am still not sure about is how do you choose which delta? For example, do i have to justify why the selected delta is correct? Can this be shown for any delta (that is constrained)? However I acknowledge that the result of the proof is incorrect. Can you explain specifically why my conclusion is wrong? Thank you for your proof but how can I fix mine? Any help appreciated – thewhale Mar 25 '24 at 17:28
  • "you probably misunderstood that" If someone misunderstands what you wrote in your proof the fault is yours for not being clear. "I was not looking to write out the entire proof with all the specific texts and exact steps I did on paper." If you don't put any effort into being coherent and understandable, you can't be surprised when someone fails to understand you. – fleablood Mar 25 '24 at 19:08
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You can prove something a bit stronger with less work.

$f(x)=\sqrt{x}$

$\sqrt{x}-\sqrt{c}= \frac{x-c}{\sqrt{x}+\sqrt{c}}$

WLOG, $x>c$

$x,c>1\implies \sqrt{x}-\sqrt{c}<(x-c)/2$

$|x-c|<\epsilon\implies \sqrt{x}-\sqrt{c}<\epsilon/2$

So $\delta=\epsilon$ proves uniform continuity in the region in question $(1,\infty)$.

Uniform continuity implies continuity and $c=2$ is in that region where the function is uniformly continuous, so the function is continuous at $x=2$.

TurlocTheRed
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