How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$? I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is there a way to rigorously show that the limit does not exist?
Note. $[[x]]$ is the greatest integer less than or equal to $x$ for all $x \in \mathbf{R}$.

Find the limits or state that they do not exist.


Problem 1.1.1 (b) $\lim_{x \to 0}x\cdot[[\frac{1}{x}]]$

Proof.
Consider the sequences $a_n:=\frac{1}{n}$ and $b_n=-\frac{1}{n}$. Both these sequences converge to zero.
The corresponding image sequences are,
\begin{align*}
(f(a_n)) &= \frac{1}{1}\cdot 1, \frac{1}{2} \cdot 2, \frac{1}{3}\cdot 3, \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
\begin{align*}
(f(b_n)) &= -\frac{1}{1}\cdot (-1), -\frac{1}{2} \cdot (-2), -\frac{1}{3}\cdot (-3), \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
Also, consider the sequence $c_n:=\frac{1}{\sqrt{n}}$. The sequence $(c_n)$ also converges to zero.
\begin{align*}
(f(c_n)) &= \frac{1}{\sqrt{1}}\cdot [[\sqrt{1}]], \frac{1}{\sqrt{2}} \cdot [[\sqrt{2}]], \frac{1}{\sqrt{3}}\cdot [[\sqrt{3}]], \frac{1}{\sqrt{4}} \cdot [[\sqrt{4}]], \ldots \\
&= 1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},1,\frac{2}{\sqrt{5}},\frac{2}{\sqrt{6}},\frac{2}{\sqrt{7}},\frac{2}{\sqrt{8}},1,\ldots
\end{align*}
Thus, the image sequence oscillates between $0$ and $1$ and is not convergent. This violates the definition of functional limits. We require that, for all sequences $(x_n)$ in the domain of the function $f$, such that $(x_n) \to a$, $x_n \ne a$ for all $n \in \mathbf{N}$, the image sequence $f(x_n)$ converges to $L$. Then, $\lim_{x \to a}f(x) = L$. So, the above limit does not exist.

Hint: $\frac 1x-1\lt [[\frac 1x]]\le 1/x $ for all $x\ne 0$
By Squeeze theorem, $\lim_{x\to 0^+} x[[\frac 1x]]=1=\lim_{x\to 0^-} x[[\frac 1x]]$
