# 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: $$0 a$$, but I've got absolutelly nothing on the third case.

• Since you are interested in the limit as $x\rightarrow 0^+$, you don't need to bother with $x\geq a$. All that matters is the behaviour of $f(x)$ for positive $x$ near $0$. Commented Sep 24, 2021 at 2:53
• @MichaelHartley thank you so much!! you're definetly my hero.
– geep
Commented Sep 24, 2021 at 3:02
• A simple observation is that if $0<x<a$ then the function value is $0$. Hence the limit is... Commented Sep 24, 2021 at 3:30

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 $$0 whilst $$\{x\}>0$$ so you have in essence: $$\lim_{y\to0}\frac{0}{0+y}$$
When you apply the limit the numerator is absolute $$0$$, but the denominator is approaching to $$0$$. Hence the limit will be $$0$$.
• Yes, I got the same result when I considered $0 <x <a$. But tks, tho!!