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Formal accepted definition: For all $\epsilon >0$, there exist $\delta >0$ such that if $0<\vert x-a\vert<\delta$, then $\vert f(x)-L\vert<\epsilon$.

The other statement: For all $\delta >0$, there exist $\epsilon >0$ such that if $0<\vert x-a\vert<\delta$, then $\vert f(x)-L\vert<\epsilon$.

I know that in quantifier logic, switching quantifiers might change the statement itself, but both of the above statements seem non-contradicting to me so must it be the same?

Blue
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TheLast Cipher
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    In the second statement $\epsilon$ can be as large as you wish, so how can you claim $f(x)$ to be close to $L$? – A.Γ. Jun 11 '18 at 03:55
  • what do you mean close to $L$? – TheLast Cipher Jun 11 '18 at 03:57
  • But doesn't the statement claim that we can pick $\epsilon$ to be not that large so it satisfies $\vert f(x)-L\vert <\epsilon$? – TheLast Cipher Jun 11 '18 at 04:01
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    The standard (first) definition of limit is a formal way to say that informally "$f(x)$ is arbitrarily close to $L$ if $x$ is sufficiently close to $a$". The second definition has no control over size of $f(x)$ missing the part that $\epsilon$ is arbitrary. – A.Γ. Jun 11 '18 at 04:01
  • Also: https://math.stackexchange.com/questions/1792551/changing-the-order-of-the-logical-symbols-forall-epsilon-exists-delta-b – amWhy Dec 29 '23 at 19:21

2 Answers2

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Consider the function $f : \mathbb{R} \to \mathbb{R}$ given by $$ f(x) = \begin{cases} 0, & x \leq 0; \\1, & x > 0. \end{cases} $$ The limit of $f(x)$ as $x$ tends to $0$ is not defined, or in other words, it is not true that for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L| < \epsilon$ for all $x$ such that $0 < |x| < \delta$. In particular, take $\epsilon = 1/2$. No matter what $L$ you choose, you cannot find such a $\delta$ (show this!).

On the other hand, look at what happens when you switch the epsilons and deltas. It is certainly true that for every $\delta > 0$ you can find an $\epsilon > 0$ such that $|f(x) - L| < \epsilon$ for all $0 < |x| < \delta$, and it doesn't even matter what $L$ you pick! Why? Let's say $L = 0$. Observe that taking $\epsilon = 2$ works (any $\epsilon > 1$ works, in fact). If you pick some other $L$, say $L = 3.5$, then choose $\epsilon = 4$. See how taking larger and larger values of $\epsilon$ allows me to make this statement true for any value of $L$ and any value of $\delta$, for this function $f$. Do you see why this fails to tell me anything about the existence of a limit? (Hint: in the second definition we are not talking about the behaviour of $f$ close to $L$. Also see @A.Γ.'s comments above.)

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An example: let $a \in \mathbb R, \delta >0, I:=(a- \delta,a+ \delta)$ and let $f:I \to \mathbb R$ be a function. Then we have:

$f$ satisfies the second definition $ \iff f$ is bounded.

Fred
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