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.