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

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.... Commented Sep 25, 2013 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? Commented Sep 25, 2013 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.... Commented Sep 25, 2013 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)$. Commented Sep 25, 2013 at 20:55 • Does this answer your question? How to find limit of function:$\lim_{x\to 0}\left(x{{\left\lfloor{ \frac{1}{x}} \right\rfloor}}\right)$Commented Dec 22, 2021 at 5:40 ## 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. Commented Sep 25, 2013 at 20:36 • @some1.new4u But this is exactly your squeeze theorem proof: $$x( \frac{1}{x} -1) \leq x \lfloor \frac{1}{x} \rfloor \leq x \frac{1}{x} \Rightarrow 1-x \leq x \lfloor \frac{1}{x} \rfloor \leq 1 \Rightarrow -x \leq x \lfloor \frac{1}{x} \rfloor -1 \leq 0 \Rightarrow \left| x \lfloor \frac{1}{x} \rfloor -1 \right| \leq |x|$$ ;) Commented Sep 25, 2013 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? Commented Sep 25, 2013 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? Commented Sep 25, 2013 at 20:57 • One more comment, I am not criticizing the answer, it is very cute and (+1). I really like the answer. ;) . Commented Sep 25, 2013 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. Commented Sep 25, 2013 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 ;) Commented Sep 25, 2013 at 21:19
• Okay. You're right. Thanks anyway. Commented Sep 25, 2013 at 21:21