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