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If $$\lim_{x \to x_0} f(x) = L$$ The $\epsilon-\delta$ definition of limit states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

If lesser and lesser values of $\epsilon$ are taken, is it possible for the maximum possible $\delta$ to increase?

i.e.: If $\epsilon=\epsilon_1$, $\delta=\delta_1$ is the maximum value that satisfies the condition, and if $\epsilon=\epsilon_2$, $\delta=\delta_2$ is the respective maximum value that satisfies the condition.

Then if $\epsilon_2<\epsilon_1$, is it possible for $\delta_2>\delta_1$?

If it is, can you give an example.

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  • $\begingroup$ @NoahSchweber I quickly realized my error, apologies $\endgroup$ Mar 18 '21 at 1:37
  • $\begingroup$ There may be no “largest value” that works. For instance, for the constant function $f(x)=L$, any $\delta$ works for every $\epsilon$. There is no “largest” $\delta$, so there is no “maximum value that satisfies the condition”. $\endgroup$ Mar 18 '21 at 1:38
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    $\begingroup$ @ArturoMagidin Sure, but that's sort of a side issue; "morally" the answer is no, that cannot happen. $\endgroup$ Mar 18 '21 at 1:39
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    $\begingroup$ @NoahSchweber: You can certainly say, as you do, that the set of valid $\delta$s does not grow; but it’s important to note that just because you talk about “the largest $X$”, that doesn’t mean such an animal exists. Its existence must be established. A better way to talk about this would be to just talk about $\Delta(\epsilon)=\{\delta\gt 0\mid \delta\text{ “works” for }\epsilon\}$, and compare the sets via inclusion. $\endgroup$ Mar 18 '21 at 1:41
  • $\begingroup$ No. If a larger $\delta_1$ assures our result, then a smaller $\delta_0$ will also assure our result because $|x-a| < \delta_0 \implies |x-a| < \delta_1 \implies |x-a| < \epsilon_1$. So the idea that a smaller $\epsilon$ might require a larger $\delta$ is simply silly. ... when taking arbitrarily small values a requirement that a value must have a larger value is completely contradictory to a value being arbitrarily small. $\endgroup$
    – fleablood
    Mar 18 '21 at 1:49
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If a given $\delta$ "works for $\epsilon_1$" and $\epsilon_2>\epsilon_1$ then that $\delta$ also "works for $\epsilon_2$" (since everything within $\epsilon_1$ of $L$ is also within $\epsilon_2$ of $L$). Making $\epsilon$ smaller can only every shrink the set of "working $\delta$s."

So there is no sense in which this can happen. Of course, a given $\epsilon$ may not have a "largest working $\delta$" at all - think about a constant function - but if we assume that such maximal $\delta$s always exist, or allow $\delta=\infty$, then your question makes sense and the situation you ask about cannot occur.

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If something is true for all $x: 0< |x-a| < \delta_1$ then it will also be true for all $x: 0< |x-a| < \delta_0 < \delta_1$ because if $0< |x-a| < \delta_0$ then it is also true that $0 < |x-a| < \delta_1$.

So there is no way that trying to fix it for a smaller $\epsilon$ would require a larger $\delta$. To require a greater $\delta$ would be to say the result fails for smaller values, and we must have the result pass for all smaller values.

Now it is possible that the $\delta_1$ required for a smaller $\epsilon$ doesn't have to be any smaller than the $\delta_0$... but it's illogical for it to be required to be larger.

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