Proving continuity with epsilon delta

Question

I have the function $f:\mathbb{R}\rightarrow \mathbb{R}\:\:f\left(x\right)=x^2-3x$ and it asks me to prove continuity in point $\:x_o=0$ using the epsilon-delta definition. I know that in order to do so I must find that for any $\epsilon >0$, there exists a $\delta >0$ so that for any $x\in \mathbb{R}$ with $\left|x-x_o\right|<\delta $ we have $\left|f\left(x\right)-f\left(x_o\right)\right|<\epsilon \:$

So now I'd have that for any $\epsilon >0$, choosing $\delta =min\left\{2,\frac{\epsilon }{7}\right\}$, we have that $\delta\gt 0$ and that for any $x\in \mathbb{R}$ with $\left|x\right|<\delta $ we have: $$\left|f\left(x\right)-f\left(x_o\right)\right|<7\left|x\right|<7\delta \le \epsilon $$

Did I do this right? I still don't understand much from it though.

Edit:

Alright how about this. We see that $\left|f\left(x\right)-f\left(0\right)\right|=\left|f\left(x\right)\right|=\left|x\right|\cdot \left|x-3\right|$ and we know $\left|x\right|<\delta $ Then we can say that $$\left|x-3\right|\le 3+\left|x\right|<\delta +3$$ So we'd have $$\left|f\left(x\right)\right|<\delta \left(\delta +3\right)$$ And the equation $t\left(t+3\right)<\epsilon $ has a solution when $t\in \left(\frac{-3-\sqrt{9+4\epsilon }}{2},\frac{-3+\sqrt{9+4\epsilon }}{2}\right)$ So then for any real $\epsilon\gt 0 $ we'd have a real $$\delta =\frac{-3+\sqrt{9+4\epsilon }}{2}$$ so that for any real x with $\left|x\right|<\delta $ we have $\left|f\left(x\right)-f\left(0\right)\right|=\left|x\right|\cdot \left|x-3\right|<\delta \left(\delta +3\right)<\epsilon $

Your calculation doesn't make sense as $|f(x) - f(0)| = |x^2 - 0| = |x|^2$ and not $|x - 3| \cdot |x|$. Did you miswrite the function? — levap, Feb 03 '16 at 12:15
Intuitively the definition is just saying that you can force the values of the function to be close $f(x_0)$ by forcing the values of the variable to be close $x_0$. That's what continuity means, things that are close are taken to things that are close - in other words there are no breaks in the function where you have points on the $x$-axis that are arbitrarily close to each other yet $f$ takes them to values that never get closer to each other than some fixed distance. — Gregory Grant, Feb 03 '16 at 12:26
@MikhaelM The problem with your argument is at first you assume $x\in(2,4)$ but later you need $|x|<\delta$. Those two conditions are incompatible. — Gregory Grant, Feb 03 '16 at 12:53
For some intuition behind the definition of continuity I invite you to read http://math.stackexchange.com/a/850218/21820 as well as the answer by user160584. — user21820, Feb 03 '16 at 13:15
@GregoryGrant I edited my question and I tried to solve the exercise a bit differently(my book provided two methods of solving such exercises). Is this by any chance right? — MikhaelM, Feb 03 '16 at 13:22
@MikhaelM: Never use the quadratic equation to solve such questions, as it provides absolutely zero understanding and is an ad-hoc method that works by luck if at all. Your earlier solution is more or less correct though it can be presented in a much clearer form. You must ask yourself why you choose such a $δ$. Well because you see when $x$ is close to zero you expect the difference $f(x)-f(0)$ to be close to zero too if $f$ is indeed continuous at $0$, and you observe that the difference factorizes into $x(x-3)$. Well $x \to 0$ and $(x-3)$ doesn't actually matter as long as it's bounded. — user21820, Feb 03 '16 at 13:32
So you choose some arbitrary interval around $0$ to make sure $(x-3)$ is bounded. You apparently chose $x \in (-1,1)$ so that $(x-3) \in (-4,-2)$, but you wrote some false line at the start instead. After that you need to see how small $x$ needs to be to make sure that $x(x-3) < ε$. I don't know why you chose $7$ but okay it works... — user21820, Feb 03 '16 at 13:35

score 1 · Accepted Answer · edited Feb 03 '16 at 13:37

1

The definition is saying this:

$f(x)$ is continuous at $x_0$ iff getting sufficiently close to $x_0$ ($|x_0 -x| < \delta$) gets us arbitrarily close to $f(x_0)$ ($\Rightarrow |f(x_0) - f(x)| < \epsilon$).

A hint for epsilon delta proofs - don't set the value of delta in the beginning, just assume $|x-x_0|<\delta$, look how $|f(x)-f(x_0)|$ will look like in terms of delta and then reverse engineer delta to be some handy function of epsilon. Then set delta to be just that (on top of you page) and look like a genius :).

Now a formal proof of continuity will go somewhat like this:

Let $\epsilon > 0$ arbitrary. If $\epsilon < 18$ take $\delta = \frac{\epsilon}{6}$. Else take $\delta = \sqrt{\frac{1}{2}\epsilon}$. Note that if $\epsilon < 18$ we have $0< \delta < 3$ and else $\delta > 3$.

Now for all x with $|x-x_0| = |x| < \delta$ we have:

$|f(x)-f(x_0)| = |x^2 - 3x| \leq |x^2|+|3x| < \delta^2 + 3\delta$

(at this point I'll start working out what delta should be in relation to epsilon)

Now depending on the case we have:

$\delta^2 + 3\delta \leq 6\delta = \epsilon$ or

$\delta^2 + 3\delta \leq 2\delta^2 = \epsilon$.

And thus $|x-x_0| < \delta \Rightarrow |f(x)-f(x_0)| < \epsilon$ meaning f is continuous in $x_0$ q.e.d.

edited Feb 03 '16 at 13:37

user21820

60,745

answered Feb 03 '16 at 12:44

Piotr Benedysiuk

1,682

Ugh, thank you for the answer, but I miswrote the function at first. I edited it to the right function now. – MikhaelM Feb 03 '16 at 12:46
Oh I see. Note this: if f(x) and g(x) are continuous then f(x) + g(x) is also continuous. So if you can prove $x^2$ continuous (which I did), you are left with proving continuity of $3x$ and that one I'll leave to you. – Piotr Benedysiuk Feb 03 '16 at 12:51
@PiotrBenedysiuk The OP may be required to do it directly from the $\delta$-$\epsilon$ definition without using the sum theorem, at least that's what it sounds like. – Gregory Grant Feb 03 '16 at 12:54
You shouldn't say "randomly", as it has the wrong meaning. – user21820 Feb 03 '16 at 12:58
@GregoryGrant Good point, working on a proof then. – Piotr Benedysiuk Feb 03 '16 at 13:05
@user21820 Thanks, I'm not used to doing math in English. What is a proper term then? – Piotr Benedysiuk Feb 03 '16 at 13:05
Well "random" always comes with a "distribution". You want to say "Let $ε$ be any given positive real." Also you didn't prove $δ > 0$ which the asker did. – user21820 Feb 03 '16 at 13:09
You can also say "Let $\epsilon>0$ be arbitrary." – Cameron Buie Feb 03 '16 at 13:16
1

@CameronBuie: I know that's the usual textbook jargon. But it hurts my ears. Let's hear it in English: "Let epsilon greater than zero be arbitrary." – user21820 Feb 03 '16 at 13:26
Okay, this should work now. $\delta > 0$ is implied by $\epsilon >0$ so I won't bother with it. – Piotr Benedysiuk Feb 03 '16 at 13:32
1

@user21820: Personally, I prefer "Take an arbitrary $\epsilon>0.$" I was just illustrating a more appropriate word than "random." – Cameron Buie Feb 03 '16 at 18:39

score 0 · Answer 2 · answered Feb 03 '16 at 12:31

0

You can take $\delta=\sqrt{\epsilon}$, then if $|x-0|<\delta\implies|x|<\delta=\sqrt{\epsilon}\implies|x|^2<\epsilon$

Now $$|x|^2<\epsilon\implies|x^2|<\epsilon\implies|x^2-0|<\epsilon\implies|x^2-0^2|<\epsilon\implies|f(x)-f(0)|<\epsilon$$

answered Feb 03 '16 at 12:31

AsdrubalBeltran

4,222

Proving continuity with epsilon delta

Edit:

2 Answers2