# Use $\epsilon-\delta$ definition of limit to prove that $\displaystyle \lim_{x \to 0} x \lfloor \frac{1}{x} \rfloor = 1$.

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.

• 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.... – N. S. Sep 25 '13 at 20:36
• @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? – user66733 Sep 25 '13 at 20:42
• 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.... – N. S. Sep 25 '13 at 20:51
• 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)$. – N. S. Sep 25 '13 at 20:55 • I added an answer explaining in detail exactly what I meant by that comment. Let me know when I should delete it. – N. S. Sep 25 '13 at 21:11 ## 2 Answers Hint: $$\left|x \left\lfloor \frac{1}{x} \right\rfloor - 1\right| = \left|x\left(\left\lfloor \frac{1}{x} \right\rfloor - \frac1x\right)\right| \le |x| = |x-0|$$ • (+1) Thanks. This is the kind of tricks I was looking for. – user66733 Sep 25 '13 at 20:36 • @some1.new4u But this is exactly your squeeze theorem proof: $$x( \frac{1}{x} -1) \leq x \lfloor \frac{1}{x} \rfloor \leq x \frac{1}{x} \Rightarrow 1-x \leq x \lfloor \frac{1}{x} \rfloor \leq 1 \Rightarrow -x \leq x \lfloor \frac{1}{x} \rfloor -1 \leq 0 \Rightarrow \left| x \lfloor \frac{1}{x} \rfloor -1 \right| \leq |x|$$ ;) – N. S. Sep 25 '13 at 20:45 • @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? – user66733 Sep 25 '13 at 20:53 • @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? – user66733 Sep 25 '13 at 20:57 • One more comment, I am not criticizing the answer, it is very cute and (+1). I really like the answer. ;) . – N. S. Sep 25 '13 at 21:12 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.

• This is an awesome answer, please don't delete it. I'm sure it will be useful to others if they read this question. – user66733 Sep 25 '13 at 21:17
• @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 ;) – N. S. Sep 25 '13 at 21:19
• Okay. You're right. Thanks anyway. – user66733 Sep 25 '13 at 21:21