Flaw in a logic reasoning about limits

Question

I have a question about an exercise from a multiple-choice test from the subject Analysis I, in the first year of Mathematics degree. The question (translated from Spanish) goes like this:

Let $f: \Bbb R \to \Bbb R$ be a function and $a \in \Bbb R, l \in \Bbb R$ such that $\lim_{x \to a} f(x) \neq l$. Then:

(i) [statement1]

(ii) [statement2]

(iii) [statement3]

It then gave possible answers like "all of them are true", "only (i) is true" etc. The correct one according to the test was "none of them is true", and I think that (i) should be true.

(i) stated this:

For some $\delta \gt 0$, there exists $\varepsilon \gt 0$ and $x \in (a-\varepsilon,a+ \varepsilon), x \neq a$ such that $\vert f(x)-l \vert \geq \delta$.

This is very similar to the negation of having limit $l$ in a ((1) from now on), but with some modifications. First, here $\varepsilon$ and $\delta$ play the opposite role of the one the usually have in a conventional definition. This is just for making it a little bit more confusing. Second, and the important one, when we negate the property of having limit $l$ in $a$, we should have "For some $\delta \gt 0$ and for all $\varepsilon \gt 0$..." instead of "For some $\delta \gt 0$, there exist $\varepsilon \gt 0$...", according to the use of letters in (i).

Now, I discussed with some other students, and a lot of them say that what the answer wanted us to say is that (i) is not equivalent to (1), but (i) is so similar to (1) that it looks that you wanted us to think that (i) is not what it should be, so (i) is false. I disagree though, because the similarities that (i) and (1) are "just coincidences", so the truth value of (i) should be given with independence of it being equivalent or not to (1). If we look at it like this, then, as (1) is given, in particular, it happens for, $\varepsilon=1$, for example. Then (i) is true, because there exist $\delta, \varepsilon, x$ which satisfy it.

I think that the reasoning involved for saying (i) is false is completely wrong, and it should be no way to justify it. Anyways, other students say that it should be understood that the question wants us to decide wether (i) is equivalent to (1), because both are very similar, and I've also asked my professor by email, and he has said that he didn't understand the question, and we will discuss it in class.

I added all the context because it think it might be reasonable to assume that the exercise wants something that isn't explicitly written, as it is a multiple-choice test.

Am I crazy or (i) is a true statement, and I'm interpreting it correctly? Is it ambiguous?

Answer 1

At least the way you worded it, it is clear that (i) is true. Pick $x_0\neq a$ such that $f(x_0)\neq\ell$ and then $\delta=|f(x_0)-\ell|$ and $\varepsilon$ large enough so that $x_0$ lies in $(a-\varepsilon,a+\varepsilon)$. It is not the negation of the statement $\displaystyle\lim_{x\to a}f(x)=\ell$, for that would be (after interchanging $\varepsilon$ and $\delta$), $$\text{for some }\delta>0, \text{for all }\varepsilon>0, \text{if }x\in(a-\varepsilon,a+\varepsilon) \text{ with }x\neq a, \text{ then }|f(x)-\ell|\geq\delta$$ This is very clearly different from the statement given. However, due to what I said in the first paragraph, the statement is a complicated way of writing $f(x)\neq\ell$ for some $x\neq a$. Rewrite this as $$\text{for some }x\neq a, \text{ we have }|f(x)-\ell|>0$$ and then $$\text{for some }\varepsilon>0\text{ and }x\in(a-\varepsilon,a+\varepsilon)\text{ with }x\neq a, \text{ we have }|f(x)-\ell|>0$$ and then $$\text{for some }\delta>0,\ \text{for some }\varepsilon>0\text{ and }x\in(a-\varepsilon,a+\varepsilon)\text{ with }x\neq a, \text{ we have }|f(x)-\ell|\geq\delta$$ Hope this helps. :)

Answer 2

Pick any $\varepsilon > 0$ (arbitrarily large), any $x \neq a$ in $(a-\varepsilon, a+\varepsilon)$ with $f(x) \neq l$, and $\delta = |f(x)-l|$, and $(1)$ is true.

This is always possible unless $f(x) = l$ for all $x \neq a$, and in that case the limit would be $l$.

The whole statement $(1)$ is just supposed to confuse you.

Answer 3

[[ Commenting in Answer Box to use MathJAX .... ]]

Check https://en.wikipedia.org/wiki/Limit_of_a_function#(%CE%B5,_%CE%B4)-definition_of_limit

It is like this :
$$(\forall \epsilon) (\exists \delta) (\forall x) (x \in (p - \delta,p + \delta)) \implies \cdots$$
( We can change $\epsilon$ & $\delta$ , though that is not the Concern here )

OP Question has that $l$ is not the limit & we can negate the limit formulation to a true statement.
That will involve something like this :
$$\lnot [[ (\forall \epsilon) (\exists \delta) (\forall x) (x \in (p - \delta,p + \delta)) \implies \cdots ]]$$
Negating the Quantifiers , we will get :
$$(\exists \epsilon) (\forall \delta) (\exists x) \cdots$$
Variable names were changed , hence it will be :
$$(\exists \delta) (\forall \epsilon) (\exists x) \cdots$$

OP Question Choice is :
$$(\text{ for some } \delta) (\exists \epsilon) \land x \in \cdots$$

(ISSUE 1) We can see that $\exists x$ is missing in the given Statement.
Is it $(\exists x)$ or $(\forall x)$ or what ? There is no Quantifier for $x$ here.
It is Imprecise & Incomplete Statement !
Hence it is not true.

(ISSUE 2) Even ignoring that ISSUE 1 , we see that $(\text{ for some } \delta)$ is weird :
Does it mean $(\exists \delta)$ ? Then it will not match the negation of the limit , hence it will be wrong. When we make the Negation , Quantifier Order will matter.
Does it mean $(\forall \delta)$ ? Then it should not say " for some " Inaccurately.
Hence it is Wrong & hence it is not true.

SUMMARY :
Statement is indeed not true.
It is either Imprecise or Incomplete or Inaccurately or Wrong.
Definitely not true !!

[[ Started to write Comment , which turned out to be Elaborate & long enough to be Answer ]]