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.
 A: 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 $N\in\mathbb N$:
$$\lim_{x\to N^-}\frac{[x^3]-x^3}{[x]-x}=1.$$
A: 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?
