An analytical way to find range of $f(x)=\frac{x}{\lfloor x\rfloor}$

Question

I want to find range of this function analytically . I tried to polt it and see $$R_f=(0,2)$$ and $$f(x)=\frac{x}{\lfloor x\rfloor}$$ Can you help me ?

Answer 1

put $x=n+p\\n \in \mathbb{Z}\\0\leq p<1 \\$

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

Answer 2

$$f(x)=\frac{\lfloor x\rfloor+\{x\}}{\lfloor x\rfloor}=1+\frac{\{x\}}{\lfloor x\rfloor}.$$

As the fractional part stays in range $[0,1)$ and the floor can equal $\pm1$, the range of the function is

$$1+(-1,0]\cup[0,1)=(0,2).$$