Define limit is not L using epsilon-delta

epsilon-deltalimits

Let $f(x)$ be a function, $x \in D$.
If for some $\epsilon >0$, there exists no $\delta >0$ such that for every $x \in D$, if $|x-a| < \delta$, then $|f(x) – L| < \epsilon$,
then $\lim_{x \to a} f(x) \neq L$.
This is what we use to prove that $\lim_{x \to a} f(x) \neq L$.

Is the converse of this statement also true? ie,
If $\lim_{x \to a} f(x) \neq L$, can we say that, for some $\epsilon >0$, there exists no $\delta >0$ such that for every $x \in D$, if $|x-a| < \delta$, then $|f(x) – L| < \epsilon$.

If not, could you give an example where the statement is false? What is the intuition behind this? Also, what is the definition of limit is not L?
I see a definition here, which begins, "there is some $\epsilon >0$ such that for every $\delta >0$…". Can we instead form a definition that begins, "for some $\epsilon >0$, there exists no $\delta >0$ such that…" ?

Best Answer

As I said in the comments: there is nothing here specific to convergence or limits. It is pure mathematical logic (more precisely: predicate logic or first-order logic).

First of all, De Morgan's laws state that: $\lnot(\forall x)P(x)$ is equivalent to $(\exists x)\lnot P(x)$, and $\lnot(\exists x)P(x)$ is equivalent to $(\forall x)\lnot P(x)$.

This is also valid if we use the constrained quantifiers such as $(\forall x>0)P(x)$. In fact, $(\forall x: P(x))Q(x)$ is just a shorthand for writing $(\forall x)(P(x)\implies Q(x))$ and $(\exists x: P(X))Q(x)$ is a shorthand for writing $(\exists x)(P(x)\land Q(x))$. Thus, for example, the following sentences are equivalent:

$$\begin{array}{c} \lnot(\forall x: P(x))Q(x)\\\lnot (\forall x)(P(x)\implies Q(x))\\(\exists x)\lnot(P(x)\implies Q(x))\\(\exists x)(P(x)\land\lnot Q(x))\\(\exists x: P(x))\lnot Q(x)\end{array}$$

and, similarly, $\lnot(\exists x: P(x))Q(x)$ can be shown to be equivalent to $(\forall x: P(x))\lnot Q(x)$.

Now, the definition of "$f(x)$ converges to $L$ when $x\to a$" (i.e. of $\lim_{x\to a}f(x)=L$) is:

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x: x\in D, |x-a|<\delta)|f(x)-L|<\epsilon$$

We are interested in how we can negate this sentence.

(Note that I would refrain from writing $\lim_{x\to a}f(x)\ne L$ because such writing is ambiguous: it is not clear whether it means

  • "When $x\to a$, $f(x)$ converges to something that is not $L$"

or it means

  • "When $x\to a$, $f(x)$ does not converge to $L$".

The latter is the negation we are looking for here; the former is a different statement, in general not equivalent to the latter.)

So, applying De Morgan's laws (three times in a row), the negation is:

$$(\exists\epsilon>0)(\forall\delta>0)(\exists x: x\in D, |x-a|<\delta)|f(x)-L|\ge\epsilon$$

(also turning $<$ into $\ge$ in the last bit). If you want to translate this to English, you would say:

There exists $\epsilon>0$ such that, for all $\delta>0$ there is an $x\in D$ such that $|x-a|<\delta$ but $|f(x)-L|\ge \epsilon$.

However, applying De Morgan's laws partially would produce other equivalent sentences, such as the following (applying the laws once):

$$(\exists\epsilon>0)\lnot(\exists\delta>0)(\forall x: x\in D, |x-a|<\delta)|f(x)-L|<\epsilon$$

There exists $\epsilon>0$ such that there is no $\delta>0$ with the property that, for all $x\in D$ such that $|x-a|<\delta$ we have $|f(x)-L|<\epsilon$"

(had to insert "with the property" to make it sound a bit nicer, but this is essentially your example from the question).

Exercise Try to apply De Morgan's laws twice. What sentence do you get, and how would you translate it to English?

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