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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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