Limit of floor function

Question

I would like to calculate these limits

$\lim\limits_{x \to 0^{+}} \frac{x}{a}{\big\lfloor\frac{b}{x}\big\rfloor}$

$\lim\limits_{x \to 0^{+}} \frac{b}{x}{\big\lfloor\frac{x}{a}\big\rfloor}$, where $a,b >0$

The answer of the first limit is $\frac{b}{a}$?

I don't know how calculate the second!

Thanks

Answer 1

Your answer for the first part is right, although it might be good to see why. The point, in broad strokes, is that when $x$ is tiny, $b/x$ is very large, and the ratio of the floor function to the argument $b/x$ itself becomes arbitrarily close to 1.

For the second question, observe that when $x < a$ (as it will be sufficiently near to zero) $$ \left\lfloor\frac{x}{a} \right\rfloor = 0$$ exactly, not approximately, so no matter what you are multiplying by, the answer will be zero.

Answer 2

Write $\;x=\lfloor x\rfloor +\{x\}\;,\;\;\{x\}=$ fractional part of $\;x\;$ , so

$$\frac xa\left\lfloor\frac bx\right\rfloor=\frac xa\left(\frac bx-\left\{\frac bx\right\}\right)=\frac ba-\frac xa\left\{\frac bx\right\}\xrightarrow[x\to0^+]{}\frac ba$$

since $\;\frac xa\xrightarrow[x\to0^+]{}0\;$ and that fractional part is bounded . Try now to argue similarly for the other limit..