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)$\lim\limits_{x \to 0} x\times\lfloor \frac{1}{x} \rfloor$
We designate t = $\frac{1}{x}$
so when x -> $0^{+}$, t -> $\infty$
so our limit becomes:
$\lim\limits_{t \to \infty}\frac{⌊t⌋}{t}$ for x -> $0^{+}$
By definition of the greatest Integer function, we know that the range of the function
is $\frac{t-1}{t}\leq\frac{⌊t⌋}{t} \leq \frac{t}{t}$
Now we take the limit of all the terms in this inequality as t->$\infty$
$\lim\limits_{t \to \infty}\frac{t-1}{t}\leq\lim\limits_{t \to \infty}\frac{⌊t⌋}{t}\leq\lim\limits_{t \to \infty}\frac{t}{t}$
We calculate the limit of the left and right sides of this inequality.
1$\leq\lim\limits_{t \to \infty}\frac{⌊t⌋}{t}\leq1$
Now, if you were like me you will probably be like "Oh! That's the squeeze theorem", then you should probably go back and do more exercises on this topic
So now we know that
$\lim\limits_{t \to \infty}\frac{⌊t⌋}{t} = 1$ and so,
$\lim\limits_{x \to 0^{+}}x\times⌊\frac{1}{x}⌋ = 1$
Now we just repeat the same steps we took for x -> $0^{-}$
$\lim\limits_{t \to -\infty}\frac{t-1}{t}\geq\lim\limits_{t \to -\infty}\frac{⌊t⌋}{t}\geq\lim\limits_{t \to -\infty}\frac{t}{t}$
$1\geq\lim\limits_{t \to -\infty}\frac{⌊t⌋}{t}\geq1$
$\lim\limits_{t \to -\infty}\frac{⌊t⌋}{t} = 1$
$\lim\limits_{x \to 0^{-}}x\times⌊\frac{1}{x}⌋ = 1$
Since the right and left-hand limits are the same we reach the conclusion:
$\lim\limits_{x \to 0} x\times\lfloor \frac{1}{x} \rfloor= 1$
