# What's wrong with this definition of limit of a function?

The standard definition of a limit of a function is the following.

Suppose $$S\subseteq \mathbb R$$, $$f:S\to \mathbb R$$ and $$x_0$$ is a limit point of S.

f has a limit L as $$x\to x_0$$ if $$\forall \epsilon>0\,\exists\delta>0$$ so that if $$|x-x_0| < \delta$$ then $$|f(x) - L| < \epsilon$$.

What's wrong with switching $$|x-x_o|$$ with $$|f(x)-L|$$ to get:

f has a limit L as $$x\to x_0$$ if $$\forall \epsilon>0\,\exists\delta>0$$ so that if $$|f(x)-L| < \delta$$ then $$|x-x_0| < \epsilon$$?

• Note the if in the expression. In that way, it implies that if a function and a number come arbitrary close, then $x$ and $x_0$ come arbitrary close. But that isn't very true. – Rebellos Mar 15 '19 at 23:43

For instance, you get that $$f(x) = 0$$ is noncontinuous which is intuitively wrong.
The fundamental idea of continuous functions is that you can predict value of function, given enough values at points that are close. So you want to tell, that when you're closer than $$\delta$$ from some $$x_0$$, then for sure, the result won't be very far away. If you reverse the relation, suddenly you lose every meaning - because you generally can't tell anything about closeness of arguments when values are the same. Just consider $$f=\sin$$. Can you tell me approximately where are its zeros, knowing that $$f(0)=0$$? I bet this is insufficient.