Prove that $\lim_{x \to 0+}\lfloor{\frac{x}{a}\rfloor}\frac{b}{x}=0$ So, I gotta prove that $\lim_{x \to 0+}\lfloor{\frac{x}{a}\rfloor}\frac{b}{x}=0$, but my only info is that $a,b \in \mathbb{R_{+}^{*}}$.
I assumed $\lfloor{\frac{x}{a}\rfloor}=\frac{x}{a}-[\frac{x}{a}]$, where $[\frac{x}{a}]$ is the fractionary part of the number $\frac{x}{a}$. So, I've got $\lim_{x \to 0+}\lfloor{\frac{x}{a}\rfloor}\frac{b}{x}=\frac{b}{a}-b\cdot \lim_{x \to 0+} \frac{[\frac{x}{a}]}{x}$. But I'm having a hard time trying to prove that $ \lim_{x \to 0+} \frac{[\frac{x}{a}]}{x}=\frac{1}{a}.$ Can you guys give me a hint?
I've found something simmilar in Show that $\lim_{x \to 0+}\frac{x}{a}\lfloor{\frac{b}{x}\rfloor}$ but $\lim_{x \to 0+}\lfloor{\frac{x}{a}\rfloor}\frac{b}{x}=0$ but it didn't help that much. I've tryied to divide into cases: $0<x<a, x=a, x> a$, but I've got absolutelly nothing on the third case.

lets say:
$$f(x)=\left\lfloor\frac xa\right\rfloor\frac bx$$
then:
$$f(x/a)=ab\frac{\lfloor x\rfloor}{x}$$
so lets just look at the function:
$$g(x)=\frac{\lfloor x\rfloor}{x}=\frac{\lfloor x\rfloor}{\lfloor x\rfloor+\{x\}}$$
notice that for $0<x<1, \lfloor x\rfloor=0$ whilst $\{x\}>0$ so you have in essence:
$$\lim_{y\to0}\frac{0}{0+y}$$
