In my other question here:

$$\lim_{x \rightarrow 9} \sqrt{x} = 3$$

We wanted to pick $\delta$ to satisfy $|\sqrt{x} - 3| < \epsilon$ where $0 < |x - 9| < \delta$

We can manipulate the delta equation to get $0 < |\sqrt{x} - 3| < \frac{\delta }{\sqrt{x} + 3} \leq \frac{\delta}{3}$ so we say $\delta = 3\epsilon$ is a valid choice of delta.

However, what about this instead:

$|\sqrt{x} - 3| < \epsilon$

$3-\epsilon < \sqrt{x} < 3 + \epsilon$

$(3-\epsilon)^2 < x < (3 + \epsilon)^2$

$(3-\epsilon)^2 - 9 < x - 9 < (3 + \epsilon)^2 - 9$

$|x - 9| < \min(9 - (3-\epsilon)^2 , (3 + \epsilon)^2 - 9)$

$|x - 9| < 9 - (3-\epsilon)^2$

$|x - 9| < 6 \epsilon -\epsilon^2$

Does this suggest that $\delta = 6 \epsilon -\epsilon^2$ is also a valid choice of delta? Is this the largest/widest delta bound I can use to satisfy the epsilon constraint?

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    $\begingroup$ when you squaring the term $3-\epsilon$ may be negative, so you should consider separate cases: $\epsilon< 3$ and $\epsilon\geq 3$. $\endgroup$ – daulomb Jan 23 '18 at 22:46
  • $\begingroup$ @daulomb If you can find a $\delta $ for $\epsilon_1 < 3$ then you can use the same $\delta$ for $\epsilon_2 > 3$. If $|x-9| < \delta \implies |\sqrt[3]x -3| < \epsilon_1$ then $|x-9| < \delta \implies |\sqrt[3]x -3| < \epsilon_1< 3 < \epsilon 2$. So we are allowed to say "As $\epsilon $ may be arbitrarily small we may assume $\epsilon < 3$". But, yes, such a consideration should be noted. $\endgroup$ – fleablood Jan 23 '18 at 22:57
  • $\begingroup$ Yes, $6e = e^2$ is a valid choice and, yes, it is the largest but... Finding the largest isn't always the point. $\endgroup$ – fleablood Jan 23 '18 at 23:01
  • $\begingroup$ Since you need to say that "for any epsilon there is a delta greater than 0...." if epsilon is too big, somewhat paradoxically, there is no delta, which means you need a little more detail. But the largest delta isn't as important as being able to say there is a delta. $\endgroup$ – Doug M Jan 23 '18 at 23:02
  • $\begingroup$ @DougM What's the better way to phrase it? $\endgroup$ – AJJ Jan 23 '18 at 23:02

This is basically correct, though to get an actual proof you would need to flesh out the logical relationships between all your inequalities to show that $|x-9|<6\epsilon-\epsilon^2$ really does imply $|\sqrt{x}-3|<\epsilon$. Note though that to prove the limit you need $\delta>0$, so this only works as a value of $\delta$ if $\epsilon<6$.

As long as $\epsilon\leq 3$, this is indeed the largest $\delta$ you can use. You can't use any larger $\delta$, since if you did, then $x=(3-\epsilon)^2$ would satisfy $0<|x-9|<\delta$ but $|\sqrt{x}-3|=\epsilon\not<\epsilon$. (The assumption that $\epsilon\leq 3$ is needed here so that $\sqrt{x}=3-\epsilon$ rather than $\epsilon-3$. If $\epsilon>3$, then it is actually impossible to have $\sqrt{x}\leq3-\epsilon$, and so the best possible $\delta$ will be $\delta=(3+\epsilon)^2-9=6\epsilon+\epsilon^2$.)

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Your method essentially tries to find all the values of $x$ for which $|\sqrt{x} - 3|<\epsilon $. This is so much different from the theme of limit definition. The idea is not to solve inequalities, but rather ensure that certain inequalities hold by constraining the values of the variable in a specific manner. The inequalities may hold even for those values of the variable which don't meet the specific constraint, but that is beside the point. What really matters is that the values of the variable which meet the specific constraints do ensure that desired inequalities hold.

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