Floor function right hand limit Show 
$$\lim_{x\to 0^+}\frac{x}a\cdot\left\lfloor\frac{b}x\right\rfloor=\frac{b}a\;.$$
I think I should use boundedness of $\left\lfloor\dfrac{b}x\right\rfloor$, $$\frac{b}x-1\le\left\lfloor\frac{b}x\right\rfloor\le\frac{b}x\;.$$
But I am not sure how to proceed with this. 

As you mention, $|\lfloor b/x \rfloor - b/x| \leq 1$, and so
$$
\frac{x}{a} \left|\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{x}\right| \leq \frac{x}{a}.
$$
We can simplify this to
$$
\left|\frac{x}{a}\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{a}\right| \leq \frac{x}{a}.
$$
As $x\to0^+$, the right-hand side tends to zero, and so $(x/a)\lfloor b/x \rfloor \to b/a$.
