# Limit involving floor function

I've been asked (by a person not by a homework) about how to compute the following limit:

$$\lim_{x \to 10^-} \frac{[x^3] - x^3}{[x] - x}$$

where $[\cdot]$ is used to denote the floor function:

$$[x] := \begin{cases} x && x \in \mathbb{Z} \\ \text{biggest integer smaller than }x && \text{otherwise} \end{cases}$$

My first thought was to sandwich this but using $x^3 - 1 \leq [x^3] \leq x^3$ to get $-1 \leq [x^3] - x^3 \leq 0$ leaves me with

$$\lim_{x \to 10^-} \frac{1}{x - [x]} \leq \lim_{x \to 10^-} \frac{[x^3] - x^3}{[x] - x} \leq 0$$

Which doesn't seem to lead anywhere. What's the right way to compute this limit? Thanks for your help.

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Since we are approaching 10 from below, just replace [x^3] with 999 and [x] with 9. The expression evaluates to 1. –  Dan Brumleve Jan 24 '12 at 11:25
@DanBrumleve: $\frac{999-1000}{9-10}$? How did you realize? I'd replace x with $9.9$ while $x$ approaches 10 from below. –  Gigili Jan 24 '12 at 12:16
@Gigili, I replaced the floor of $x$ with 9 because it is the greatest integer less than 10, and likewise for $x^3$ and $999$. –  Dan Brumleve Jan 24 '12 at 12:18
@DanBrumleve: Right, sounds reasonable. –  Gigili Jan 24 '12 at 12:21
@Gigili, I'm really not certain so I didn't answer. :) –  Dan Brumleve Jan 24 '12 at 12:23

We have, putting $x=N+\delta$: $$\lim_{x\to N^-}\frac{[x^3]-x^3}{[x]-x}=\lim_{\delta\to 0^-}\frac{[(N+\delta)^3]-(N+\delta)^3}{[N+\delta]-(N+\delta)}.$$ We have $[N+\delta]-N=N+[\delta]-N=[\delta]=-1$ when for example $-\frac 12\leq\delta<0$, and $$[(N+\delta)^3]-(N+\delta)^3=[N^3+3N^2\delta+3N\delta^2+\delta^3]-(N^3+3N^2\delta+3N\delta^2+\delta^3)=[a(\delta)]-a(\delta),$$ where $a(\delta)=3N^2\delta+3N\delta^2+\delta^3$. Since $a(\delta)<0$ when $\delta<0$ and when $|\delta|$ is small enough, it's $\geq -1$ so we conclude that for any $N\in\mathbb N$: $$\lim_{x\to N^-}\frac{[x^3]-x^3}{[x]-x}=1.$$

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Hmm, there are still the $\small \delta^2$ - terms wrong/missing, and also in the second fraction's denominator you've replaced x by N instead of $\small N+\delta$ However, I arrived at the same result, too. –  Gottfried Helms Jan 24 '12 at 13:29
@GottfriedHelms Indeed, there are $\delta^2$ terms (and with copy-pasting I made the mistake several times). THe other missing $\delta$ has been added. –  Davide Giraudo Jan 24 '12 at 13:34
Yes, this "writing-in-the-box" is often a hazzle (and for longer sequences also I employ an external editor and copy&paste and thus many additional sources-of-errors...) :-) –  Gottfried Helms Jan 24 '12 at 13:54

Different from Davide's I replace $\small x=N-\delta$ where $\small \delta \to 0$. Also I would change sign in numerator and denominator to subtract the smaller from the larger values:

$\qquad \small { (N-\delta)^3 - [(N-\delta)^3 ] \over (N-\delta) - [N-\delta]}$

which for small enough $\delta$ and N=10 comes to

$\qquad \small { (10- \delta)^3 - [1000- \epsilon ] \over (10- \delta) - 9} = { (1000-300\delta+30\delta^2-\delta^3) - 999 \over 1-\delta} = { 1-300\delta+30\delta^2-\delta^3 \over 1-\delta}$

Here we insert $\small 0= -\delta + \delta$ and get

$\qquad \small { 1-\delta + \delta - 300\delta+30\delta^2-\delta^3 \over 1-\delta} =1 + \delta \cdot { 1 - 300+30\delta-\delta^2 \over 1-\delta} =1 - \delta \cdot { 299 -30\delta +\delta^2 \over 1-\delta}$

whose limit is 1 if $\small \delta \to 0$ , the same what Davide already got.

[update] Hmm, after a second read I question, whether we can talk of a "limit" here because if we approximate from $\small N+\delta$ we arrive at +300 where we have $\small {0 \over 0}$ when $\small x=10$ So perhaps this can answer/comment another one for the correct terminology?

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