# Negation of the Definition of Limit of a Function

Question

Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

and show this by the symbolism

$$\mathop {\lim f(x)}\limits_{x \to a} = L$$

Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from

$$\neg \left[ {\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)} \right]$$

and then go through it to get the final form of negation (See the example below).

My Thought

I just wrote down the two following negations without going through a step by step approach.

$$\forall L,\exists \varepsilon > 0:\left( {\forall \delta > 0,\exists x:0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$

or

$$\nexists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

I want to know weather these are true or not.

Example

I will give an example of what I mean by a step by step approach. Consider the following statement

$${P}\implies R$$

I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic

\begin{align} \, \neg \left( {P \implies R} \right) &\iff \neg \left( {\neg P \vee R} \right) & \text{Conditional Disjunction} \\ \qquad \qquad \quad &\iff \neg \neg P \wedge \neg R & \text{Demorgan's Law} \\ \qquad \qquad \quad &\iff P \wedge \neg R & \text{Double Negation} \end{align}

• Try to negate the following first: $\forall L \exists M\colon L<M$. You might already spot a mistake in the above. Oct 10, 2015 at 11:07
• I am asking if there is a mistake! :) Please spot any mistakes if it exists. Thanks. Oct 10, 2015 at 11:10
• It is a bit non-standard to write $\forall x \in A \rightarrow P$ where $A$ is a set and $P$ is a formula. Oct 10, 2015 at 11:54
• From FAQ about tags: Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag. It's not easy to keep balance between too specific tags and not having enough tags, but it is always good to search first and to ask yourself, whether newly created tag is not too specific. (Of course, you can disagree with the removal of the tag you've created, and there is possibility for further discussion, if needed.) Oct 14, 2015 at 5:13

The first negation is almost completely right. You forgot to negate the implication at the end.

Remember that the negation of an "if, then" statement is not an "if, then" statement. $A \implies B$ has negation "$A \land \neg B$" (read: $A$ and not $B$).

So, "if the sky is blue, then I love cheese" has negation "the sky is blue and I do not love cheese."

We say $\lim \limits_{x \to a} f(x) = L$ if $$\forall \epsilon > 0\text{, }\exists \delta > 0 \text{ such that }\forall x\text{, }|x - a| < \delta \implies |f(x) - L| < \epsilon.$$ Then the negation of this is: $$\exists \epsilon > 0\text{ such that }\forall \delta > 0\text{, }\exists x\text{ such that }|x - a| < \delta \text{ **and** }|f(x) - L| \geq \epsilon.$$

UPDATE Here is how to negate the following statement step by step

Negation of "$\lim \limits_{x \to a} f(x)$ exists", i.e., $$\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| \geq \epsilon).$$

We say $\lim \limits_{x \to a} f(x)$ does not exist if:

$\neg[\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L \neg[\forall\epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\neg[\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\forall \delta > 0\neg[\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: \neg[|x - a| < \delta \implies |f(x) - L| < \epsilon]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land \neg(|f(x) - L| < \epsilon)$ (Negation of implication)

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$

• @H.R. I'm afraid I can't help because even though I know how to negate it step by step, but I am unfamiliar with the names of the tautologies I would be using. If you would like to see the negation step by step anyway, let me know and I will show you (no need to award the bounty to me as I would be giving you only part of what you want). Oct 12, 2015 at 21:33
• Of course I am interested. Please edit your answer in this regard. :) Oct 13, 2015 at 8:20
• Comments are not for extended discussion; this conversation has been moved to chat. Oct 13, 2015 at 8:45
• @H.R. I updated the answer. Is that what you are looking for? We should discuss this in the chat set up by Daniel Fischer, but I was commenting to send you a notification. Oct 13, 2015 at 11:12

Your two negations are true and exactly the same. I prefer this:

$$\forall L\;\exists \varepsilon > 0:\left(\forall \delta > 0\to\left\{ {x|\;0 < \left| {x - a} \right| < \delta }\;and\; \left| {f(x) - L} \right| \ge \varepsilon\right\}\neq\emptyset\right)$$

• Nice simple negation form. There is also another interesting question.How do you prefer to write the definition? Oct 10, 2015 at 13:37
• Hello. I think your answer is wrong in terms of the OP's negations being true. The OP's negations are not correct. The negation of an "if, then" statement isn't an "if, then" statement. Oct 10, 2015 at 15:37
• I have passed no course in mathematical logic but think it's an equivalent one at least for this statement. Oct 10, 2015 at 18:54
• H. R, $$\mathop {\lim f(x)}\limits_{x \to a} = L$$ iff $$\forall \varepsilon > 0\;\exists \delta > 0:\left( 0 < \left| {x - a} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon\right)$$ Oct 10, 2015 at 19:26
• @A.F.23 Before "$0 < |x - a| < \delta$" you should have "$\forall x \text{ such that}$". Also, again, your sentence of "Your two negations are true and exactly the same" is false. Oct 11, 2015 at 0:45

If non exist $\lim_{x \to x_0}f (x)$ then for all $L$, $\exists \epsilon > 0$ such that for all $\delta > 0 ,\exists x : |x-x_0|<\delta \Longrightarrow |f(x)-L|>\epsilon.$

• Isn't this just the first negation, I wrote above with some defects? :) I think you must add that there exist an $x$ belonging to that neighborhood not for all $x$. Also you forgot the equality sign. Oct 10, 2015 at 11:20
• No. It's good :) Oct 10, 2015 at 11:22
• You adjoint: forall x in the negation :) Oct 10, 2015 at 11:25
• I didn't get you. Please write down your answer in my notation (don't use any words for negation) and tell me where I was wrong. Oct 10, 2015 at 11:26
• Still not good. You must also negate $$|x-x_0|<\delta\implies |f(x)-L|>\varepsilon$$
– user228113
Nov 19, 2015 at 11:39

We say $$\lim\limits_{x→a}f(x)=L$$ if

∀ϵ>0, ∃δ>0 such that ∀x, |x−a|<δ⟹|f(x)−L|<ϵ.

Then the negation of this is: ~(∀ϵ>0, ∃δ>0 such that ∀x, |x−a|<δ⟹|f(x)−L|<ϵ) ==

∃ϵ>0 such that ∀δ>0, ∃x such that |x−a|<δ ^ |f(x)−L|≥ϵ.