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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 $\epsilon$?

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    @Ben, what I'm thinking is that 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 $\epsilon$?2011-10-13

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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. $

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    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.2011-10-15