Is it possible for $\delta$ to increase when $\epsilon$ is decreasing 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.
 A: 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.
A: 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.
