To prove "\(\displaystyle \lim_{x\to 3} x^2= 9\)" you must show that "given any \(\displaystyle \epsilon> 0\) there exist a \(\displaystyle \delta> 0\) such that if \(\displaystyle |x- 3|< \delta\) then \(\displaystyle |x^2- 9|< \epsilon\)".
The best way to see that and determine what that \(\displaystyle \delta\) must be for a given \(\displaystyle \epsilon\) is to work backwards: we want to go from \(\displaystyle |x^2- 9|= |(x- 3)(x+ 3)|= |x- 3||x+ 3|< \epsilon\) to \(\displaystyle |x- 3|< \delta\). Well, you can see that we already have the "|x- 3|". What do we do with "|x+ 3|"?
Since we want to make |x- 3| small anyway, start by requiring that |x- 3|< 1 (1 is pretty much chosen arbitrarily- any positive number would work). That is the same as saying -1< x- 3< 1. Adding 6 to each part, 5< x+ 3< 7. Since -7< 5, we can write -7< x+ 3< 7 so that |x+ 3|< 7.
Now, we can write that |x- 3||x+ 3|< 7|x- 3| so if we make \(\displaystyle 7|x- 3|< \epsilon\) it will certainly be true that \(\displaystyle |x- 3||x+ 3|< \epsilon\). And \(\displaystyle 7|x- 3|< \epsilon\) is the same as \(\displaystyle |x- 3|< \frac{\epsilon}{7}\).
Keeping in mind our requirement that |x- 3|< 1, we see that we must have |x- 3| less than both 1 and \(\displaystyle \frac{\epsilon}{7}\) so we say that \(\displaystyle \delta= min(1, \frac{\epsilon}{7})\) where "min" is the minimum or lesser of the two.
Now, the actual "proof" or the limit (which, having shown that we can find \(\displaystyle \delta\), we seldom write out!) would go like this: given \(\displaystyle \epsilon> 0\) take \(\displaystyle \delta\) to be the smaller of 1 and \(\displaystyle \epsilon/7\) so that we have both \(\displaystyle |x- 3|< 1\) and \(\displaystyle |x- 3|< \epsilon/7\). Then we certainly have \(\displaystyle |x- 3|< \epsilon/7\) so that \(\displaystyle 7|x- 3|< \epsilon\). But since, as shown above, if \(\displaystyle |x- 3|< 1\), \(\displaystyle |x+ 3|< 7\), \(\displaystyle |x^3- 9|= |x- 3||x+ 3|< 7|x- 3|< \epsilon\) and we are done.
