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I recently proved the following statement:

$$\text{ if } \displaystyle \lim_{x \to a^-}f(x)=M \text{ and } \displaystyle \lim_{x \to a^+}f(x)=N \text{ where } M\neq N \text{ then} \lim_{x \to a}f(x) \text{ Does Not Exist }$$

I carried out this proof by demonstrating the truth of the following FOL statement:

$$\forall L \in \mathbb R \ \exists \varepsilon \gt 0 \ \forall \delta \gt 0 \ \exists x \in \mathbb R\ \left [0 \lt |x-a| \lt \delta \land|f(x)-L| \geq \varepsilon \right ]$$

After finishing this proof, I wondered if the work I spent was rather wasteful. I knew from previous exercises that the following biconditional was true:

$$\displaystyle \lim_{x\to a}f(x)=L \leftrightarrow \displaystyle \lim_{x \to a^-}f(x)=L \text{ and }\displaystyle \lim_{x \to a^+}f(x)=L$$

I figured because it was a biconditional, I could have taken the negations of each side, where one negation implies the other. However, I then ran into some difficulties trying to determine what the negations of the left and right statements of the biconditional are.

I came up with two possibilities:

  1. $\displaystyle \lim_{x\to a}f(x)\neq L \leftrightarrow \displaystyle \lim_{x \to a^-}f(x)\neq L \text{ or }\displaystyle \lim_{x \to a^+}f(x) \neq L$

  2. $\displaystyle \lim_{x\to a}f(x) \text { DNE} \leftrightarrow \displaystyle \lim_{x \to a^-}f(x) \text { DNE } \text{ or }\displaystyle \lim_{x \to a^+}f(x) \text { DNE}$

I quickly ruled out 2. because by my proof above, it is clearly not the case the that the "DNE" of a limit at $a$ implies the "DNE" of the limit at $a^-$ or $a^+$.

So that means it must be 1. However, the "$\displaystyle \lim_{x\to a}f(x)\neq L$" portion of 1. goes against my intuition regarding what $\neg \left [\displaystyle \lim_{x\to a}f(x)=L \right]$ equals. Is it (perhaps) valid to claim that $\displaystyle \lim_{x\to a}f(x) \text { DNE} \rightarrow \displaystyle \lim_{x\to a}f(x)\neq L$?

A few months ago, I posted a question using the logic tag that asked about the difference between existence and equality in the context of limits (Question about the difference between equality and existence in the context of limits.). I was unable to arrive at any particular resolution regarding this distinction. If anyone could clear up my confusion, I would greatly appreciate it!

S.C.
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    As to the one question you asked in the post, whether it is "valid to claim that $\lim_{x\to a}f(x) \text { DNE} \rightarrow \lim_{x\to a}f(x)\neq L$": yes, that is a valid implication (though its converse is not valid). Indeed, the definition of the left-hand side is $\forall Y\in\Bbb R, \lim_{x\to a}f(x)\ne Y$, and so this implication is just applying the definition to $Y=L$. – Greg Martin Jul 21 '21 at 17:51
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    $$\lim_{x \rightarrow a} f(x) = L $$ is an abbreviation for a slightly more complex statement. Since you tagged your question with real-analysis, it is equivalent to either $$\forall {x_n \rightarrow a} \lim_{n\rightarrow \infty} f(x_n) = a $$ or equivalently

    $$\forall \varepsilon > 0 \exists \delta >0 \forall x: |x-a| <\delta \Rightarrow |f(x)-f(a)| < \varepsilon$$

    Try negating any of these statements to see what's wrong with your reasoning.

     – Thomas Jul 21 '21 at 17:57
  • @Thomas ahhh. I see my error. I (incorrectly) assumed that $\exists L \in \mathbb R \ \forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ \forall x \in \mathbb R\ \left [0 \lt |x-a| \lt \delta \rightarrow |f(x)-L| \lt \varepsilon \right ]$ was logically equivalent to $\forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ \forall x \in \mathbb R\ \left [0 \lt |x-a| \lt \delta \rightarrow |f(x)-L| \lt \varepsilon \right ]$. I see now that while $\displaystyle \lim_{x\to a}f(x)=L$ implies the existence of some $L$, the converse is not true (be wary of the overloaded semantics behind the symbol $L$) – S.C. Jul 21 '21 at 20:21

1 Answers1

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The negation of $$\lim_{x \to a}f(x) = L$$ is $$\lim_{x \to a}f(x) \text{ DNE or } \lim_{x \to a}f(x) \neq L.$$ So the negated biconditional reads $$\lim_{x \to a}f(x) \text{ DNE or } \lim_{x \to a}f(x) \neq L \iff \lim_{x \to a^{-}}f(x) \text{ DNE or } \lim_{x \to a^{-}}f(x) \neq L \text{ or } \lim_{x \to a^+}f(x) \text{ DNE or } \lim_{x \to a^+}f(x) \neq L.$$

But to prove your statement all you need is the fact that if $\lim_{x \to a}f(x)$ exists, then $\lim_{x \to a^-}f(x) = \lim_{x \to a^+}f(x) = \lim_{x \to a}f(x)$.

Mason
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  • For your final statement, you are suggesting to take the contrapositive, yes? – S.C. Jul 21 '21 at 18:40
  • @S.Cramer I was thinking by contradiction. If $\lim_{x \to a}f(x)$ exists, it's impossible for the left and right limits to have different values. – Mason Jul 21 '21 at 18:45