what is the limit of $\displaystyle \lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$

Question

find the limit of:

$\displaystyle\lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$

$ a,b>0$

i know that: $\lfloor\frac{b}{x}\rfloor\le \frac{b}{x} + 1$; and $\lfloor\frac{b}{x}\rfloor\ge \frac{b}{x} -1$

but i cant seem to combine those using the squeezing theorem (i belive the limit is $\frac{b}{a}$)

Answer 1

$$\lfloor{\frac bx\rfloor}=\frac bx-\left\{\frac bx\right\}\implies$$

$$\frac xa\lfloor\frac bx\rfloor=\frac xa\frac bx-\frac xa\left\{\frac bx\right\}\xrightarrow[x\to 0]{}\frac ba$$

Since $\;\frac xa\to 0\;$ and $\;\{b/x\}\;$ is bounded...

Answer 2

Consider two cases: $x>0$ and $x<0$. In each case multiply your inequalities by $\frac{x}{a}$. Then, say for $x<0$, you get:

$\frac{b}{a} -\frac{x}{a} \geq \frac{x}{a} \lfloor \frac{b}{x} \rfloor \geq \frac{b}{a}+ \frac{x}{a}.$

Now use the squeeze theorem.