# Implication in the $(ε,δ)$-definition of limit

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 $x_0$?

• Suppose for instance that $f$ is constant (with value $l$). Dec 16, 2017 at 19:08

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:

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.

• So basically my definition does not work, except for injective functions. Am I right? Dec 16, 2017 at 19:47
• @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 $2$. Dec 16, 2017 at 19:58
• If i restrict $x^2$ to positive reals, doesn't it become injective in the positive reals? Dec 16, 2017 at 20:00
• @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. Dec 16, 2017 at 20:00
• @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. Dec 16, 2017 at 20:11

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 $|x-x_0|<\delta$.

• 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? Dec 16, 2017 at 19:23
• @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. Dec 16, 2017 at 19:35
• Why can't we derive ε given a certain δ? Dec 16, 2017 at 20:54
• @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 $\delta$. Dec 16, 2017 at 21:33
• @Davide La Vardera, it's because the definition says $\forall\varepsilon\,\exists\delta$ and not $\forall\delta\,\exists\varepsilon$. Dec 16, 2017 at 21:49

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.