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

Question

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 $x_0$?

Answer 1

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.

Answer 2

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.

Answer 3

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 $|x-x_0|<\delta$.