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 $?

  • $\begingroup$ You should find any $\delta(\varepsilon) \gt 0$ for which holds inequalities. So if you find specific $\delta(\varepsilon)$, then it works also. $\endgroup$
    – zkutch
    Commented Apr 7, 2021 at 10:04
  • 5
    $\begingroup$ $\delta $ is not even a uniquely determinde function of $\epsilon$ in the definition. So increasing, decreasing etc don't make sense. $\endgroup$ Commented Apr 7, 2021 at 10:04
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2021 at 10:07

1 Answer 1


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.

If there is no reason to enforce a condition, do not enforce it, that could make some proofs more complicated than necessary.

  • $\begingroup$ 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 $\endgroup$
    – mathslover
    Commented Apr 19, 2021 at 20:47
  • 1
    $\begingroup$ @mathslover: if $\delta_1$ works, anything smaller works. $\endgroup$
    – user65203
    Commented Apr 20, 2021 at 6:27
  • $\begingroup$ @YvesDaoust Could you provide a reason as to why anything smaller to $\delta_{1}$ works? $\endgroup$
    – Nameless
    Commented Apr 27, 2021 at 0:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .