# 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.

• @NoahSchweber I quickly realized my error, apologies Mar 18 '21 at 1:37
• 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”. Mar 18 '21 at 1:38
• @ArturoMagidin Sure, but that's sort of a side issue; "morally" the answer is no, that cannot happen. Mar 18 '21 at 1:39
• @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. Mar 18 '21 at 1:41
• 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. Mar 18 '21 at 1:49

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.
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.