Find $ \lim_{x\to a^-}(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor $ This problem is an updated version of this given here and I found it interesting for myself I would like no if the following limit exists or not.
$$ \lim_{x\to a^-}(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor $$
Here the interesting fact is that we have the singularity at $x =a$ which not the case here
Any idea,?
 A: Introducing a new variable $y=\frac{1}{x-a}$ allows us to rewrite this as
$$\lim_{y\to-\infty} \frac{\lfloor y\rfloor}{y}$$
now using the fact that $$y-1\leq \lfloor y\rfloor \leq y$$
gives us $$\frac{y-1}{y}\geq \frac{\lfloor y\rfloor}{y}\geq \frac yy$$
(the inequalities reverse when dividing by a negative number) simplifying to
$$1-\frac1y \geq \frac{\lfloor y\rfloor}{y}\geq1$$
which means the limit is $1$ by the squeeze theorem.

You can also do it without the change of variables, since $$\frac{1}{x-a}-1\leq \left\lfloor\frac{1}{x-a}\right\rfloor\leq \frac{1}{x-a}$$ and multiplying by $(x-a)$ gives
$$1-(x-a)\geq (x-a)\left\lfloor\frac{1}{x-a}\right\rfloor \geq 1$$ and again, the limit is $1$.
A: For $a - \frac{1}{n} < x < a - \frac{1}{n+1}$, we have $-\frac{1}{n} < x - a < - \frac{1}{n+1}$. Hence, $-(n+1) < (x-a)^{-1} < -n$ and $\lfloor (x-a)^{-1} \rfloor = -(n+1)$. Therefore,
$$ 1 < (x-a) \lfloor \frac{1}{x-a} \rfloor < \frac{n+1}{n}.$$
Your desired limit is equivalent to taking the limit as $n \to \infty$, so
$$ \lim_{x \to a} (x-a) \lfloor \frac{1}{x-a} \rfloor = 1.$$
A: WLOG, $a=0$ (otherwise, shift the variable).
$$ \lim_{x\to 0}x\left\lfloor \frac{1}x\right\rfloor=\lim_{y\to\pm\infty}\frac {\lfloor y\rfloor}{y}=1-\lim_{y\to\pm\infty}\frac {\{y\}}y=1$$
as the fractional part is bounded. (Note that the limit works on both sides.)

A: We know that
$$\left\lfloor \frac{1}{x-a}\right\rfloor\le \frac{1}{x-a}<\left\lfloor \frac{1}{x-a}\right\rfloor+1$$
so that for $a>x$ we have
$$(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor\ge1 >(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor+(x-a)$$
which equivalent to 
$$0\ge1 -(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor>(x-a)$$
Now taking the limite we have 
 $$\color{red}{\lim_{x \to a^-} (x-a)\left\lfloor \frac{1}{x-a}\right\rfloor = 1}$$
