Alternative Definition of Formal Limit

Question

After encountering the formal definition of limit over and over again, I wondered why it has been defined in such a way rather than something else. For instance, if we swap $0 < |x - c| < \delta$ with $|f(x) - L| < \epsilon$ to get:

$$\lim_{x \to c} f(x) = L \iff [\forall \epsilon \in \mathbb{R}^{+}\ \exists \delta \in \mathbb{R}^{+}\ \forall x \in \mathbb{R}\ (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta)] \tag{1}$$

From the original formal definition of limit in symbolic logic shown below:

$$\lim_{x \to c} f(x) = L \iff [\forall \epsilon \in \mathbb{R}^{+}\ \exists \delta \in \mathbb{R}^{+}\ \forall x \in \mathbb{R}\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)] \tag{2}$$

I know the two are not logically equivalent statements. Thinking in logic terms let $P(\delta, x)$ and $Q(\epsilon ,f(x))$ designate open sentences $0 < |x - c| < \delta$, $|f(x) - L| < \epsilon$, respectively, so that we have a general truth-value table for every $\epsilon$, $\delta$ and $x$ interpretation throughout their domain of discourse:

$$\begin{array}{|c|c|c|c|} \hline P(\delta, x) & Q(\epsilon ,f(x)) & P(\delta, x) \implies Q(\epsilon ,f(x)) & Q(\epsilon ,f(x)) \implies P(\delta, x)\\ \hline True & True & True & True\\ \hline False & True & True & False\\ \hline True & False & False & True\\ \hline False & False & True & True\\ \hline \tag{3} \end{array}$$

If we move on with an example, say $f(x) = 2$, we want to find its limit at $1$ ($\lim_{x \to 1} f(x) = 2$) using the two definitions above. For a given $\epsilon > 0$ we can rewrite for both definitions:

$$P(\delta, x) : 0 < |x - c| < \delta \rightarrow \fbox{$0 < |x - 1| < \delta$}\\ Q(\epsilon ,f(x)) : |f(x) - L| < \epsilon \rightarrow |2 - 2| < \epsilon \rightarrow \fbox{$0 < \epsilon$}$$

Here I claim that the truth value of $Q(\epsilon ,f(x))$, which is reduced to $0 < \epsilon$ as shown above, is $True$ independent of any $x$ and $\delta$ values. That is, the truth value of $Q(\epsilon ,f(x))$ is always $True$ for all $x$ and $\delta$, also depends only on the domain of discourse of $\epsilon$, which is defined above as $\forall \epsilon \in \mathbb{R}^{+}$. So we can eliminate the last two rows of table $(3)$. Also by definition the truth value of $\lim_{x \to 1} f(x) = 2$ is $True$ for both definitions $(1)$ and $(2)$ above. Therefore the truth value of the entire right-hand side of the original definition $(2)$ after $\iff$ symbol becomes trivially $True$ as $P(\delta, x) \implies True$ pattern fits into the first two rows of table $(3)$ under $P(\delta, x) \implies Q(\epsilon ,f(x))$ column. So the left and right-hand side of $\iff$ for $(2)$ are both $True$.

However, if we apply the second definition $(1)$ on aforementioned constant function in a similar way, we get $True \implies P(\delta, x)$, whose truth values falls under $Q(\epsilon ,f(x)) \implies P(\delta, x)$ column in the first two rows. Therefore we get $False$ for

$$[\forall \epsilon \in \mathbb{R}^{+}\ \exists \delta \in \mathbb{R}^{+}\ \forall x \in \mathbb{R}\ (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta)]$$

of $(1)$ so that $\lim_{x \to 1} f(x) \neq 2$, which is absurd as we know a constant function has limit at all points of its domain.

Therefore we stick with the original definition. I am also aware of the fact that what I have tried above is in a sense tautological kind of thinking, which makes no sense.

Does such a treatment make sense in terms of logic to show why mathematical definitions, theorems, etc. are the way we use? I now asked such a question even though the noteworthy replies to my pre-edited version of the same question helped me grasp why not use such a definition to some extent, but I still feel a bit unsure of what I took away from them. That is why I further elaborated my thinking above as succinctly as possible.

Answer 1

You make two proposals, but do not give them names, so I've decided to give them random names and not tell you which is which.

The new definition, Fred, has an immediate negative consequence: $$ \lim_{x \rightarrow 0} \sin x $$ does not exist. Here, we have $L = c = 0$. For any $\varepsilon > 0$, there is a neighborhood of $k \pi$, for every integer $k$, on which $|\sin(x) - 0| < \varepsilon$. But there is no $\delta$ so large that $|k \pi - 0| < \delta$ for all integers $k$.

Your definition requires that all points having the same height be nearby in the $x$ direction.

You new definition, Bob, is only satisfied by constant functions.

Neither of these captures the notion of a limit.

Answer 2

Your definition does not correspond to continuity. Informally, it says "If $f(c)$ is sufficiently close to $f(x)$ then $c$ is close to $x$." But this clearly doesn't hold of even very nice continuous functions - for example, take any constant function.

In fact, it's a good exercise to show that if $f$ is not injective then $f$ fails to have your property for some $x$. Since continuity doesn't really have anything to do with injectivity - again, constant functions are very continuous but very non-injective - your property has no connection to continuity.

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EDIT: You ask why constant functions don't satisfy your definition.

Let's consider at $f(x)=0$, at the specific point $c=0$, and look at the specific choice $\epsilon=1$. In order for $f$ to satisfy your property, we need some $\delta$ such that for all $x$ we have

if $\vert f(x)-0\vert<\epsilon$ then $x$ is within $\delta$ of $0$.

But since $f$ is constant, we have $\vert f(x)-0\vert<\epsilon$ for every $x$ whatsoever! So we would need some $\delta$ such that every $x$ is within $\delta$ of $0$, and there is no such $\delta$. This is just a fancy way of saying "There is no largest (or smallest) real number."

(What if we allow $\delta=\infty$? Well, that would solve this problem but would introduce a new one: this $\delta$ would work for every function at every point. So that doesn't fix things.)