$\forall x$ versus $\exists x$ in the definition of limit

Question

Suppose $f: A \to \Bbb R$ and $x \in A$. Let $\lim\limits_{x \to c} f(x) = L,$ that is,

$$\forall \varepsilon > 0, \exists \delta > 0 \quad \text{such that if} \quad x \in A, \quad \text{ and } |x - c| < \delta \to |f(x) - L| < \varepsilon.$$

Why is it "for all $x$" instead of "there exists an $x$"? If $f=x=8$ and $L=c=5,$ then the condition won't be satisfied since $|f - L| = 3 \not< 1 = \varepsilon$. However, writing "$\exists x$" is solving the problem, because then I can always find some $x$ that won't break the condition.

Answer 1

To have your problem solved, we need to come back to the beginning.

What you have written up there, is called Epsilon ($\epsilon$)- Delta ($\delta$) proof of limit, which is a precise way to define limits. Before you have this kind of definition, you probably use this type of definition instead:

Given a function $y = f(x)$, and an x-value, $c$, we have the limit of $f(x)$ as it approach $c$ is $L$ if one or more of these cases happen: $y$ tends to reach $L$ as $x$ tends to $c$; $y$ approaches $L$ as $x$ approaches $c$; $y$ is near $L$ whenever $x$ is near $c$.

When you define it as above, you realise, that it is not exact. In what way that those relations of "tends", "approach" or "near" the variable actually is? That is when you have the Epsilon-Delta proof. It is the result of formalizing the third statement:

If $x$ is within the tolerance level of $c$, then the corresponding value $y = f(x)$ is within the tolerance level of $L$.

From here, you can see that, we are taking the approximation that is defined as "tends to" or "approach" to the tolerance level, the degree of accuracy required in a measurement or the acceptable range of values in which the measurement will be acceptable in our calculation. Which is further reduced into notation by saying:

If $x$ is within $\delta$ unit of $c$, then the corresponding value of $y$ is within $\epsilon$ units of $L$.

Mathematically, this is written as

$$|x-c| < \delta \longrightarrow |y - L| < \epsilon$$

with the LHS as $$c - \delta < x < c + \delta$$ and the RHS as $$L - \epsilon < y < L - \epsilon$$

Formally, we would have this definition of limit

Definition: The limit of a function $f(x)$: Let $I$ be an open interval that $c,x \in I$,$x \neq c$, $f(x)$ is defined everywhere in $I$ except possibly at $c$. The function $f(x)$ is defined as $f: I \rightarrow \mathbb{F}$. The limit of $f(x)$ as $x$ approaches c is $L$, denoted by $$ \lim_{x\to c} f(x) = L $$ is such that $$\forall \: \epsilon > 0, \exists \: \delta \: s.t. \: |x-c| < \delta \longrightarrow |f(x) - L| < \epsilon $$

From all of this, I think it is suffice for you to understand why it must be that way. Another remark is that, x is independent of y. And you are misunderstanding both misreading the definition.

Reference: Epsilon-Delta definition of a limit

Answer 2

First of all, sorry for Saturday's verbosity and let's start:

The difference between the word "all" and the word "some" ("exists") is obvious: compare "all the money in the world is mine" and "some of the money in the world is mine". Maybe we would like the first, but we live in the second.

Thus, it is clear that you significantly relax the requirements if the universal quantifier is replaced by the existential quantifier.

But let's look at the issue more broadly: quantifiers are designed to impose conditions on variables, but they are, let's say, too extreme - or all or at least one. Let's consider a generalization of the limit to the case when we impose a condition on a part of the set.

Let's assume $f:E\to\mathbb{R}$, where $E\subset\mathbb{R}$ and $c$ is limit point for $E$

Formally limit for function $\lim\limits_{x\to c}f(x)=L$ with respect to $E$ is following (Vladimir A. Zorich - Mathematical Analysis I-Springer (2016), page 106 or Limit_of_a_function) :

$$(\forall \varepsilon>0)(\exists \delta>0)(\color{red}\forall x \in E)\Big(0<|x-c|<\delta\Rightarrow |f(x)-A|<\varepsilon \Big)$$

Also we can write it as following

$$(\forall \varepsilon>0)(\exists \delta>0)\Big(\color{red}\forall x \in E\cap \{x:0<|x-c|<\delta\}\Big)\Big( |f(x)-A|<\varepsilon \Big)$$

By choosing a different $E$ we can control the extent to which the second universal quantifier (indicated in red) is applied.

Let's consider the well-known Dirichlet function

$$D(x)=\begin{cases} 1, & x\in \mathbb{Q} \\ 0, & x\notin \mathbb{Q} \end{cases}$$

If we take $E= \mathbb{R}$, then limit doesn't exists, but it exists if $E= \mathbb{Q}$ or $E= \mathbb{R}\setminus\mathbb{Q}$.

Returning to your situation, limit exists also when $E= \mathbb{R}$ and we change red $\forall$ to $\exists$.