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I'm still confused by the use of  $\Rightarrow$  in (ε,δ)-definition of limit.
Take for example the definition of $\underset{x\rightarrow x_{0}}{\lim}f\left(x\right)=l$ :

$$\forall\varepsilon>0,\;\exists\delta>0\quad\mathrm{such\:that\quad}\forall x\in\mathrm{dom}\,f,\;0<\left|x-x_{0}\right|<\delta\;\Rightarrow\;\left|f\left(x\right)-l\right|<\varepsilon$$

My questions are:

Why is $\left|f\left(x\right)-l\right|<\varepsilon$ not a sufficient condition for $0<\left|x-x_{0}\right|<\delta\;$?

Or, stated in another way, shouldn't $\left|f\left(x\right)-l\right|<\varepsilon\;\Rightarrow\;0<\left|x-x_{0}\right|<\delta\;$ also be true ? If $f\left(x\right)$ becomes arbitrarily close to $l$, doesn't $x$ becomes arbitrarily close to $x_0$?

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    $\begingroup$ Suppose for instance that $f $ is constant (with value $l $). $\endgroup$
    – Andrés E. Caicedo
    Dec 16, 2017 at 19:08

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Already given example of constant function should (in my opinion) be enough to shoot the whole idea down in a blazing glory, but parabola might be more convincing visually:

enter image description here

As we can see, $\lim_{x\to -2} x^2 = \lim_{x\to 2} x^2 = 4$, and when we are getting close to the limit $4$ on the $y$-axis, it could be that we are either close to $2$ or $-2$, but most definitely we can't get close to both at the same time.

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  • $\begingroup$ So basically my definition does not work, except for injective functions. Am I right? $\endgroup$
    – Davide La Vardera
    Dec 16, 2017 at 19:47
  • $\begingroup$ @Davide La Vardera, actually, no. Restrict $x\mapsto x^2$ to positive reals and take the same limit. Then $-1 < x^2 -4 < 1 \iff \sqrt 3 - 2 < x-2 < \sqrt 5 - 2$, so such $\delta$ would exist for $\varepsilon = 1$. Notice that interval that is the preimage is not symmetric around $2$. $\endgroup$
    – Ennar
    Dec 16, 2017 at 19:58
  • $\begingroup$ If i restrict $x^2$ to positive reals, doesn't it become injective in the positive reals? $\endgroup$
    – Davide La Vardera
    Dec 16, 2017 at 20:00
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    $\begingroup$ @Davide La Vardera, of course it does, otherwise it wouldn't be counterexample to your claim. And now that I think about it, the mere condition in the definition that $0<|x-x_0|$ already breaks the whole thing. $\endgroup$
    – Ennar
    Dec 16, 2017 at 20:00
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    $\begingroup$ @Davide La Vardera, if it's not clear enough, $|f(x)-l|<\varepsilon \implies 0<|x-x_0|<\delta$ can be rephrased as $f^{-1}(l-\varepsilon,l+\varepsilon) = (x_0-\delta,x_0+\delta)\setminus \{x_0\}$ which definitely isn't true. $\endgroup$
    – Ennar
    Dec 16, 2017 at 20:11
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Changing the definition in that way would mean that a constant function cannot have a limit, for example.

Or as a less trivial example, consider for example $\lim\limits_{x\to 1}\frac1x$. Intuitively this ought to be $1$, but with your addition to the definition the limit would not exist. Namely, if I choose $\varepsilon=2$ then you can't find any $\delta$ such that $|\frac1x-1|<2$ is only true when $|x-x_0|<\delta$.

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  • $\begingroup$ I see now. Thanks to all of you for the answers. I still don't understand why in the proof of limits we assume that $\left|f\left(x\right)-l\right|<\varepsilon$ in order to derive that $0<\left|x-x_{0}\right|<\delta$. Shouldn't it be the opposite? $\endgroup$
    – Davide La Vardera
    Dec 16, 2017 at 19:23
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    $\begingroup$ @DavideLaVardera: If you want to prove that $l$ is a limit from the definition, you should assume that somebody gives you an unknown $\varepsilon$ and then you need to come up with a $\delta$ such that $0<|x-x_0|<\delta$ implies $|f(x)-l|$. You may be confused by didactic presentation that are about exploring how one might find a $\delta$ that will work -- but this exploration is not part of the proof, just an illustration of the thought process that could lead you to discover details that will allow the proof to work. $\endgroup$
    – hmakholm left over Monica
    Dec 16, 2017 at 19:35
  • $\begingroup$ Why can't we derive ε given a certain δ? $\endgroup$
    – Davide La Vardera
    Dec 16, 2017 at 20:54
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    $\begingroup$ @DavideLaVardera: The definition says that for each $\varepsilon$ there must be a qualifying $\delta$. If you start with an arbitrary $\delta$ and work your way to an $\varepsilon$ that works with that, you won't have any guarantee that all possible $\varepsilon$s get a $\delta$. $\endgroup$
    – hmakholm left over Monica
    Dec 16, 2017 at 21:33
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    $\begingroup$ @Davide La Vardera, it's because the definition says $\forall\varepsilon\,\exists\delta$ and not $\forall\delta\,\exists\varepsilon$. $\endgroup$
    – Ennar
    Dec 16, 2017 at 21:49
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You can think about it in this way:

first set $\epsilon$ and then you have to find $\delta$ such that the inequality:

$$\left|f\left(x\right)-l\right|<\varepsilon$$

is satisfied.

If you can do it for every $\epsilon>0$ then the limit exists.

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