Formal definition of limit (epsilon delta)

DEF.$$\lim_{x \rightarrow x_0} f(x)=L \Leftrightarrow \forall \varepsilon >0: \exists \delta >0: 0<\left|x-x_0 \right|< \delta \Longrightarrow \left|f(x)-L \right|<\varepsilon$$

Why doesn't the definition have any other requirements for $$\delta$$, for example that when $$\varepsilon$$ decreases, $$\delta$$ decreases as well?

If we for example found $$\delta$$ s.t. $$\delta=\delta (\varepsilon)=\begin{cases} \varepsilon+1, \ \ \ &\mathrm{if}\ 0<\varepsilon \le 2\\ \varepsilon-2, \ \ \ &\mathrm{if}\ \varepsilon > 2. \end{cases}$$

and $$\forall \varepsilon>0: 0 < \left|x-x_0 \right|<\delta (\varepsilon) \Longrightarrow \left| f(x)-L \right|<\varepsilon,$$ would it still imply that $$\lim_{x \rightarrow x_0} f(x)=L$$?

And the same question for limits at infinity:

DEF. $$\lim_{x \rightarrow \infty} f(x)=L \Leftrightarrow \forall \varepsilon>0: \exists M>0: \ x>M \Longrightarrow \left|f(x)-L \right|<\varepsilon$$

If we found $$M = M(\varepsilon)$$ s.t. M actually decreases as $$\varepsilon$$ decreases, would it still imply, that $$\lim_{x \rightarrow \infty} f(x)=L$$?

• You should find any $\delta(\varepsilon) \gt 0$ for which holds inequalities. So if you find specific $\delta(\varepsilon)$, then it works also. Commented Apr 7, 2021 at 10:04
• $\delta$ is not even a uniquely determinde function of $\epsilon$ in the definition. So increasing, decreasing etc don't make sense. Commented Apr 7, 2021 at 10:04
• If for some $\varepsilon$ there is a $\delta$ or an $M$ that works, then all positive numbers smaller than $\delta$ and all numbers greater than $M$ also work. Nowhere does it say that they have to be optimal. An optimal choice for $\delta$ might not even exist. So we can always choose $\delta(\varepsilon)$ and $M(\varepsilon)$ such that they are not monotonous. This doesn't hurt our limit definition one bit, though. Commented Apr 7, 2021 at 10:07

If some $$\delta$$ works for a given $$\epsilon$$, then all smaller $$\delta$$'s will work as well. Assume that $$\delta(\epsilon)\le\epsilon$$ works in all cases (for instance with $$f(x)=x$$), then $$\delta(\epsilon)=\epsilon \dfrac{\cos\epsilon+2}3$$ is equally valid.
• I am still having hard time understanding this intuitively, but then this idea (which probably is the same as you presented) came to me: So suppose that I find $\delta_1$ s.t. $\forall \varepsilon>0: \ \exists \delta_1>0: \ 0<\left| x-x_0 \right|<\delta_1 \Longrightarrow \left| f(x)-L \right|<\varepsilon$. Then let's choose $\delta_2 = \mathrm{min} \left\{\delta_1, \varepsilon \right\}$ which works for delta in the definition as well. Now it's clear that with small $\varepsilon \ \delta_2$ is small as well. Is this idea correct? @Yves Daoust Commented Apr 19, 2021 at 20:47
• @mathslover: if $\delta_1$ works, anything smaller works.
• @YvesDaoust Could you provide a reason as to why anything smaller to $\delta_{1}$ works? Commented Apr 27, 2021 at 0:42