Answer
(a) The limit does not exist.
(b) $\lim\limits_{x \to 0} x~\lfloor 1/x \rfloor = 1$
Work Step by Step
(a) $\lim\limits_{x \to 0^+} \frac{\lfloor x \rfloor}{x}$
$=\lim\limits_{x \to 0^+} \frac{0}{x}$
$= 0$
$\lim\limits_{x \to 0^-} \frac{\lfloor x \rfloor}{x}$
$=\lim\limits_{x \to 0^-} \frac{-1}{x}$
$= \infty$
Since the left limit and the right limit are not equal, the limit does not exist.
(b) Suppose $\frac{1}{n+1} \lt x \leq \frac{1}{n}$ for some positive integer $n$:
Then $~~n+1 \gt \frac{1}{x} \geq n~~$ and $~~\lfloor 1/x \rfloor = n$
Then $~~\frac{n}{n+1} \lt x~\lfloor 1/x \rfloor \leq \frac{n}{n}$
$\frac{n}{n+1} \lt x~\lfloor 1/x \rfloor \leq 1$
Then:
$\lim\limits_{x \to 0^+} x~\lfloor 1/x \rfloor = 1$
Suppose $-\frac{1}{n-1} \lt x \leq -\frac{1}{n}$ for some positive integer $n$:
Then $~~-(n-1) \gt \frac{1}{x} \geq -n~~$ and $~~\lfloor 1/x \rfloor = -n$
Then $~~\frac{n}{n-1} \gt x~\lfloor 1/x \rfloor \geq \frac{n}{n}$
$\frac{n}{n-1} \gt x~\lfloor 1/x \rfloor \geq 1$
Then:
$\lim\limits_{x \to 0^-} x~\lfloor 1/x \rfloor = 1$
Therefore:
$\lim\limits_{x \to 0} x~\lfloor 1/x \rfloor = 1$
