# How to express $\delta$ in an expression when $|x-3|= \frac {\epsilon}{|x+3|} = \delta$, but without using the $x$ term

if $$|x-3|= \frac {\epsilon}{|x+3|} = \delta$$

if we take a given value of $\epsilon$, then $x$ is constrained to some value by the above equation. Is there a function, lets say $g()$, that will give that $x$ value, given $\epsilon$ as the input, so that we can compute $\delta$ as $\frac {\epsilon}{g(\epsilon) + 3}$. Or, alternatively, is there some way that, knowing the above relation, we can find $\delta$ given $\epsilon$?

-
Are you tring to find $\lim_{x \rightarrow 3} f(x)$ where $f(x) = x^2$? – user38268 Oct 13 '11 at 21:51
This is very confusing to me. Firstly, you have here that $\delta$ is a function of x. And you have two functions of $x$ that are related by $\epsilon$, so $\epsilon$ is also a function of x. What is going on? – mixedmath Oct 13 '11 at 21:56
@mixedmath I am trying to figure that out myself... – user38268 Oct 13 '11 at 21:56
@Ben, yes, and I know of one way to do it, but I'm curious about this. – Matt Munson Oct 13 '11 at 21:57
Yeah I can, but in the first place your question above is very confusing. You may consider rephrasing it. – user38268 Oct 13 '11 at 22:07

Assume that $|x-3|\leqslant\delta$, then $|x+3|\leqslant6+\delta$ hence $|x^2-3^2|=|x+3|\cdot|x-3|\leqslant\delta(6+\delta)$. Now, $\delta(6+\delta)\leqslant\varepsilon$ for every $\delta\leqslant a(\varepsilon)$, where $a(t)=\sqrt{9+t}-3$ for every nonnegative $t$. This proves the following assertion: $$\forall\varepsilon>0,\quad\forall x\in\mathbb R,\quad|x-3|\leqslant a(\varepsilon)\implies |x^2-3^2|\leqslant\varepsilon.$$
Very nice. If I understand correctly, you actually show two ways to prove $\lim_ {x\rightarrow 3}$ of $x^2 = 9$. You can either set $\delta$ relative to $\epsilon$ using $\delta = \sqrt(9+\epsilon) - 3$ or you can set $\epsilon$ relative to $\delta$ using $\epsilon = \delta(6+ \delta)$. – Matt Munson Oct 14 '11 at 0:52
Hmmm... not really. What you call the second way has anything to do with continuity only because $\delta(6+\delta)\to0$ when $\delta\to0$--a fact which is quantified by the function $\delta\mapsto a(\delta)$ in what you call the first way. Hence, in the end there is only one proof of the continuity, which starts by fixing $\varepsilon$ and looks for $\delta$ such that what you know holds. – Did Oct 14 '11 at 5:14
Is it not true that $|x^2 - 3^2| < \epsilon$ as long as $\epsilon = \delta(6+\delta)$? Ie, $|(\delta+3)-3| * |(\delta+3)+3| - \delta(6+\delta) = 0$ as long as $\delta$ is non-negative. And isn't that sufficient to prove the limit, because we can see that we can make $\epsilon$ as small as we want by making $\delta$ sufficiently small? – Matt Munson Oct 14 '11 at 22:13
Matt, the whole point (once again) is to prove that we can make $\varepsilon$ as small as we want by making $\delta$ sufficiently small. How do you know this is true? Highbrow answer: because the function $u:\delta\mapsto\delta(6+\delta)$ is $0$ and continuous at $\delta=0$. Back-to-the-definitions answer: because for every $\delta$ in $[0,a(\varepsilon)]$, $u(\delta)\leqslant\varepsilon$. Both answers have their merits but I suspect that the latter is more adapted to what you know and what you ask. – Did Oct 15 '11 at 4:03