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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 $\left|f\left(x\right)-f\left(x_o\right)\right|<\epsilon \:$

This definition I know by heart, but I don't really understand it. Anyhow, following from past examples, I did this: I said that $$\left|x-3\right|<1\:for\:x\in \left(2,4\right)$$ and $$\left|x\left(x-3\right)\right|=\left|x\right|\cdot \left|x-3\right|$$ so we have $$\left|x-3\right|\le \left|x\right|+3<7$$ and this together would mean that $$\left|f\left(x\right)-f\left(0\right)\right|=\left|x-3\right|\cdot \left|x\right|\le 7\cdot \left|x\right|$$

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 $

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    $\begingroup$ 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? $\endgroup$ – levap Feb 3 '16 at 12:15
  • $\begingroup$ I want to prove it for $x_o=0$ $\endgroup$ – MikhaelM Feb 3 '16 at 12:17
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    $\begingroup$ @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. $\endgroup$ – Gregory Grant Feb 3 '16 at 12:53
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    $\begingroup$ @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. $\endgroup$ – user21820 Feb 3 '16 at 13:32
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    $\begingroup$ 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... $\endgroup$ – user21820 Feb 3 '16 at 13:35
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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.

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  • $\begingroup$ Ugh, thank you for the answer, but I miswrote the function at first. I edited it to the right function now. $\endgroup$ – MikhaelM Feb 3 '16 at 12:46
  • $\begingroup$ 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. $\endgroup$ – Piotr Benedysiuk Feb 3 '16 at 12:51
  • $\begingroup$ @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. $\endgroup$ – Gregory Grant Feb 3 '16 at 12:54
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    $\begingroup$ @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." $\endgroup$ – user21820 Feb 3 '16 at 13:26
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    $\begingroup$ @user21820: Personally, I prefer "Take an arbitrary $\epsilon>0.$" I was just illustrating a more appropriate word than "random." $\endgroup$ – Cameron Buie Feb 3 '16 at 18:39
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You can take $\delta=\sqrt{\epsilon}$, then if $|x-0|<\delta\implies|x|<\delta=\sqrt{\epsilon}\implies|x|^2<\epsilon$

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

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