put $x=n+p\\\
\in \mathbb{Z}\\\0\leq p<1 \\\$
$$\quad{f(x)=\frac{x}{\lfloor x\rfloor}=\frac{n+P}{n}=\\\1+\frac{p}{n}}$$now look for $max \frac pn$ $$case 1: n>0 \to 0<\frac pn<1\\\case 2: n<0 \to -1<\frac pn<0\\\obviously \space n\
eq 0 $$so $$0<\frac pn<1 \to 0+1<1+\frac pn<2\\\\-1<\frac pn<0 \to -1+1<1+\frac pn<0+1 \\\case 1 \cup case 2:(0,2)$$