7
$\begingroup$

Prove that $\lim_{x\to 2}x^2=4$ using $\epsilon-\delta$ definition.

By the mean of $\epsilon-\delta$ definition, $|x-2|\le \delta,|x+2|\le \delta+4$

then $|x-2||x+2|\le \delta(\delta+4),|x^2-4|\le \delta^2+4\delta$.

Assign $\epsilon=\delta^2+4\delta,|x^2-4|\le\epsilon$.

Q.E.D.

Is this method correct? If yes, why I always see people do this by putting $\delta = \min(1,\epsilon/5)$...?

Thanks.

$\endgroup$
  • 1
    $\begingroup$ You must fix $\epsilon$ at the begining, that's why you see much convolution to choose a $\delta$ that will do the job. Remember, that looks like $\forall \epsilon > 0, \exists \delta > 0, \ ...$ $\endgroup$ – Stop hurting Monica Mar 14 '13 at 13:52
  • $\begingroup$ Here is a related problem. $\endgroup$ – Mhenni Benghorbal Mar 14 '13 at 14:02
  • $\begingroup$ Then can I find the $\delta$ by solving $\delta^2+4\delta-\epsilon=0$? $\endgroup$ – A. Chu Mar 14 '13 at 14:24
  • 1
    $\begingroup$ @jasoncube: Follow the technique in the problem I referred you to it and you do not have to do all these calculations. $\endgroup$ – Mhenni Benghorbal Mar 14 '13 at 20:04
9
$\begingroup$

This is pretty good, but this goes the other way around.

You fix $\epsilon>0$ first.

Then you it suffices to find $\delta>0$ such that $\delta^2+4\delta\leq\epsilon$. You need to make this explicit. So indeed, lots of people would do it like this. If you take a priori $\delta\leq 1$, then $$ \delta^2+4\delta\leq \delta+4\delta=5\delta. $$ So it suffices to have $\delta \leq\epsilon/5$. If $\epsilon/5\leq 1$, the choice $\delta =\epsilon/5$ works by the above estimate. If not, the choice $\delta=1$ works, still by the previous estimate.

That's why most people would take $$ \delta:=\min\left(1,\frac{\epsilon}{5}\right) $$ which works either way.

$\endgroup$
3
$\begingroup$

No that one won't work, as you say $\epsilon=\dots$

The definition says that for every $\epsilon>0$ there is a delta, so you stell need to proof, that with your setting of $\epsilon$ you still can reach every value in $(0,\infty)$.

If you set $\delta $ to something you don't get those problems.

$\endgroup$
1
$\begingroup$

As pointed out by others, you have to follow the structure of the $\epsilon$-$\delta$ criterion in your proof. Try to fit your proof into the following frame:

Let $\epsilon > 0$.

We set $\delta = \ldots$ [Find some suitable $\delta > 0$ depending on $\epsilon$.]

Let $x\in\mathbb{R}$ with $\left|x - 2\right| < \delta$.

Then ...

So $\left|x^2 - 4\right| < \epsilon$.

$\endgroup$
0
$\begingroup$

We have a fixed $\varepsilon \gt 0$ and we want to set $\delta \gt 0$ so that for every $\delta_0 \in (-\delta,+\delta)\;$ (i.e. $|\delta_0| \lt \delta$),

$\tag 1 4- \varepsilon \lt 4 + 4 \delta_0 + {\delta_0}^2 \lt 4 + \varepsilon$

Now if $\delta = min(1, \frac{\varepsilon}{5})$,

$\quad 4 \delta_0 + {\delta_0}^2 \le |4 \delta_0| + |\delta_0| = 5 \, | \delta_0| \lt \varepsilon $

so the inequality on the rhs of $\text{(1)}$ will be true for this setting of $\delta$.

Fortunately, this setting also works for the lhs of $\text{(1)}$:

We have

$\quad -4 \delta_0 \le 4 |\delta_0 | \lt 4 \delta \le \frac{4 \varepsilon} {5} \lt \varepsilon$

But then $-\varepsilon \lt 4 \delta_0 \lt 4 \delta_0 + {\delta_0}^2$.

Note: For the lhs we only need to set $\delta = \frac{\varepsilon}{5}$. I guess we were lucky that we started with the rhs!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.