Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$

Hint: Suppose that $a$ is positive. Then $$\left\lfloor\frac{a}{x}\right\rfloor=\frac{a}{x}-\delta(a,x),$$ where $0\le \delta(a,x)\lt 1$. Now multiply by $x$ and take the limit.

The case $a$ negative will be very similar, and $a=0$ is obvious.

$$\frac{a}{x} - 1 \le E(\frac{a}{x}) \le \frac{a}{x}$$

$$a- x \le xE(\frac{a}{x}) \le a$$

if you apply the limit it gives $a$

Note: $E(x)$ is the step function.

$\bullet$ First show $\displaystyle \lim_{x\to \infty}\frac{[x]}{x}=1$. For this ,

$$\left|\frac{[x]}{x}-1\right|=\left|\frac{[x]-x}{x}\right|<\frac{1}{|x|}\to 0 \text{ as , } x\to \infty.$$

$\bullet$ Now put , $\frac{a}{x}=y$. Then limit becomes $\displaystyle \lim_{y\to \infty}\frac{a[y]}{y}=a$ .