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For example, consider this question-

Use the $(\epsilon, δ)$-definition to prove the existence or non-existence of the following limit- $$f : R → R, f(x) := [x]$$

$$ \lim_{x→0} f(x)$$

Here we do not know apriori if the limit exists or not. Now I am confused about whether I should start with the assumption that the limit exists or instead assume it does not exist (I mean which is an easier way to show it). Also, should I try to assume the opposite of the correct statement to be true and try to use a counterexample or should I try to show the correct statement it in a direct way?

Soham
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  • Show that $\lim_{x→0^+} f(x)$ and $\lim_{x→0^-} f(x)$ exist and they are different. – Robert Z Oct 19 '19 at 07:22
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    First off, from the graph of the function, what can you say about the limit? –  Oct 19 '19 at 07:22
  • @AnotherJohnDoe It does not exist. So do you advise to keep the real result at the back of my mind before proceeding to use epsilon-delta? – Soham Oct 19 '19 at 07:24
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    @tatan Yes, knowing (or at least strongly suspecting) what the result is will have a great effect on how you start your $\epsilon$-$\delta$ proof. – Arthur Oct 19 '19 at 07:25
  • @Arthur Okay so it is kind of cheating right? – Soham Oct 19 '19 at 07:30
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    Well, it's not like suspecting the answer and letting that guide you will let you skip any of the logic and reasoning. – Arthur Oct 19 '19 at 08:07
  • @Arthur Thanks. Makes sense :-) – Soham Oct 19 '19 at 08:08

2 Answers2

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You have to assume that the limit exists and get a contradiction. Suppose $[x] \to l$ as $ x\to 0$. Then there exits $\delta >0$ such that $|[x]-l| <\frac 1 2$ whenever $|x|<\delta$. Since we may replace $\delta$ by any smaller number we may suppose $\delta <1$. Take $x=\delta /2$ to see that $|0-l|<\frac 1 2$. Then take $x=-\delta /2$ to get $|(-1)-l| <\frac 1 2$. Combine these two to get $1<1$, a contradiction.

Kavi Rama Murthy
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  • Why is $[\frac{\delta}{2}]=0$? – Soham Oct 19 '19 at 07:29
  • I have edited my answer to impose the condition $\delta <1$. @tatan – Kavi Rama Murthy Oct 19 '19 at 07:32
  • Can we arbitrarily any value of $x$ ? I mean what is the logical path exactly like... I didn't understand what we are really doing – Soham Oct 19 '19 at 08:09
  • The definition says $\forall \epsilon$ there exists $\delta$... But here we assume one specific $\epsilon=1/2$. Why? – Soham Oct 19 '19 at 08:10
  • @tatan I am assuming that the limit exists, so for every $\epsilon >0$ some thing happens, right? So I am at freedom to use this property for $\epsilon =\frac 1 2$. If something holds for every positive number it must hold for $\frac 1 2$. – Kavi Rama Murthy Oct 19 '19 at 11:37
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A simpler way is to check by $(\epsilon, δ)$ definition that

  • $\lim_{x→0^+} f(x)=0$
  • $\lim_{x→0^-} f(x)=-1$
user
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  • @tatan The main fact is that for $|x|\le 1$ for $x>0 \implies f(x)=0$ and for $x<0 \implies f(x)=-1$. – user Oct 19 '19 at 08:33