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I was trying to write some nice problems for applying $\epsilon-\delta$ definition to give it to my friend but then I realized that I couldn't solve some of them either. This is one of them:

Use $\epsilon-\delta$ definition of limit to prove that

$$ \lim_{x \to 0} x \lfloor \frac{1}{x} \rfloor = 1$$

It's easy to show that this is true by using the squeeze(sandwich) theorem, but I'm looking for an $\epsilon-\delta$ proof.

Also, a similar problem could be:

$$ \lim_{n \to \infty} \frac{[nx]}{n}=x$$

Again it's obvious that this is true by using the squeeze theorem, but I'm looking for an elementary proof that uses nothing but just the definition of the limit of a sequence.

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  • $\begingroup$ Once you squeze it, the $\epsilon-\delta$ proof comes out from the squeeze nicely. And I am pretty sure whatever what you do, you cannot avoid squeezing, because that is what integrer part is. You might be able to hide the squeeze in a nice cute way so it doesn't look like squeeze, but it will still be squeeze.... $\endgroup$ – N. S. Sep 25 '13 at 20:36
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    $\begingroup$ @N.S. Would you explain more precisely why you say that? njguliyev's answer just uses the simple fact that $|\lfloor x \rfloor - x|<1$. Am I wrong? $\endgroup$ – user66733 Sep 25 '13 at 20:42
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    $\begingroup$ Your squeeze theorem proof is, I guess, $x(\frac{1}{x}-1) \leq x \lfloor \frac{1}{x} \rfloor \leq 1$. But this inequality is equivalent to $-x \leq x \lfloor \frac{1}{x} \rfloor -1$ which implies the inequality njg got.... $\endgroup$ – N. S. Sep 25 '13 at 20:51
  • $\begingroup$ What I am really trying to say is that there is no difference between saying that $|\lfloor \frac{1}{x} \rfloor - \frac{1}{x}|<1$ and squeezing $x \lfloor \frac{1}{x} \rfloor -1$ between $-|x|$ and $|x|$ (or 0)$. $\endgroup$ – N. S. Sep 25 '13 at 20:55
  • $\begingroup$ I added an answer explaining in detail exactly what I meant by that comment. Let me know when I should delete it. $\endgroup$ – N. S. Sep 25 '13 at 21:11
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Hint: $$ \left|x \left\lfloor \frac{1}{x} \right\rfloor - 1\right| = \left|x\left(\left\lfloor \frac{1}{x} \right\rfloor - \frac1x\right)\right| \le |x| = |x-0|$$

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  • $\begingroup$ (+1) Thanks. This is the kind of tricks I was looking for. $\endgroup$ – user66733 Sep 25 '13 at 20:36
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    $\begingroup$ @some1.new4u But this is exactly your squeeze theorem proof: $$x( \frac{1}{x} -1) \leq x \lfloor \frac{1}{x} \rfloor \leq x \frac{1}{x} \Rightarrow 1-x \leq x \lfloor \frac{1}{x} \rfloor \leq 1 \Rightarrow -x \leq x \lfloor \frac{1}{x} \rfloor -1 \leq 0 \Rightarrow \left| x \lfloor \frac{1}{x} \rfloor -1 \right| \leq |x|$$ ;) $\endgroup$ – N. S. Sep 25 '13 at 20:45
  • $\begingroup$ @N.S. Fair enough. But do you mean that this is always like that no matter what function we're dealing with or you're just referring to this particular problem? $\endgroup$ – user66733 Sep 25 '13 at 20:53
  • $\begingroup$ @N.S. Please add it as an answer. I get what you're saying about this particular problem, but my confusion is if this is true in general. I mean is it true that if we prove something with squeeze theorem we can find such an inequality or this is just a coincidence in this particular case? $\endgroup$ – user66733 Sep 25 '13 at 20:57
  • $\begingroup$ One more comment, I am not criticizing the answer, it is very cute and (+1). I really like the answer. ;) . $\endgroup$ – N. S. Sep 25 '13 at 21:12
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Sorry, to long for a comment, will delete it in the future. Here is exactly what I mean by my comment.

Assume that on some interval $I=(a-b,a+b)$ around $a$ we have. $$f(x) \leq g(x) \leq h(x) \, \forall x\in I \backslash \{ a \}$$

and the outside limits are easy, meaning that you can prove with $\epsilon-\delta$ that $\lim_{x \to a} f(x) =\lim_{x \to a} h(x)= L$. Then you get for free a proof with $\epsilon-\delta$ that $\lim_{x \to a} g(x)= L$.

Indeed

$$f(x) \leq g(x) \leq h(x) \Rightarrow f(x) -L \leq g(x)-L \leq h(x)-L \Rightarrow $$ $$\left|g(x)-L \right| \leq \max\{ \left| f(x) -L \right| , \left| h(x) -L \right| \} (*)$$

Now, pick an $\epsilon >0$, pick the corresponding $\delta_1$ for $g$ and $\delta_2$ for $h$ and set $\delta = \min \{ \delta_1, \delta_2 \}$. Thus

Then if $0 < |x-a | < \delta$ you have $0 < |x-a | < \delta_1$ and $0 < |x-a | < \delta_2$

$$ \left| f(x) -L \right| < \epsilon, \left| h(x) -L \right| < \epsilon \,,$$

and if you plug these in $(*)$ you are done.

For this problem, the heuristic reason why I think that, no matter what the approach is, if it is simple it is a hidden squeeze theorem argument: the simplest way of relating $\lfloor y \rfloor$ to $y$ for all real $y$ at once is $y-1 \leq \lfloor y \rfloor \leq y$. Moreover, the bounds can be attained, so you can't improve it. But once you do that, it becomes exactly the argument I included.

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  • $\begingroup$ This is an awesome answer, please don't delete it. I'm sure it will be useful to others if they read this question. $\endgroup$ – user66733 Sep 25 '13 at 21:17
  • $\begingroup$ @some1.new4u Ok, I will leave it but since it doesn't really answer the direct question you asked (it answers a related question) it is only fair if you accept the other answer ;) $\endgroup$ – N. S. Sep 25 '13 at 21:19
  • $\begingroup$ Okay. You're right. Thanks anyway. $\endgroup$ – user66733 Sep 25 '13 at 21:21

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