<s>[INST] what is the limit of $\displaystyle \lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$ 
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 $\frac{b}{a}$)
 [/INST] 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.
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