How can I calculate this limit: $\lim\limits_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor$? Calculate the following limit.
$$
\lim_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor
$$
Where $\left\lfloor x \right\rfloor$ represents greatest integer function or floor function, i.e greatest integer less than or equal to $x$.
Thanks.
 A: If $\frac1x=n+y$ where $n$ is any integer and $ 0\le y<1,\implies  \left[\frac1x\right]=n$
So, $x \left[\frac1x\right]=\frac n{n+y}=\frac1{1+\frac yn}$ 
As $x\to 0$ and  $ 0\le y<1, n\to\infty\implies \lim_{x\to 0}x \left[\frac1x\right]=\lim_{n\to\infty}\frac1{1+\frac yn}=1$ 
A: Consider $x \in (1/(n+1),1/n]$. We then have $\dfrac1x \in [n,n+1)$. Hence, $\left\lfloor \dfrac1x \right\rfloor = n$. Hence, we have
$$x \left\lfloor \dfrac1x \right\rfloor \in \left(\dfrac{n}{n+1},1\right]$$
Argue similarly, for $x \to 0^{-}$. Now use the above to show that
$$\lim_{x \to 0} x \left\lfloor \dfrac1x \right\rfloor = 1$$
A: If $[x]$ means floor (usually noted $\lfloor x \rfloor$), you have that $\lfloor x \rfloor = -1$ if $-1 \le x < 0$, while $\lfloor x \rfloor = 0$ when $0 \le x < 1$. The limit can't exist.
A: Use one basic inequality concerning integer part
$$\frac{1}{x}\leq [\frac{1}{x}]\leq \frac{1}{x}+1$$
\begin{align}
1 \leq x[\frac{1}{x}]\leq 1+x \,\mathrm{if}\, x>0\\
1\geq x[\frac{1}{x}]\geq 1+x \,\mathrm{if}\, x<0
\end{align}
So by squeeze theorem $$\lim _{n\rightarrow 0}x[\frac{1}{x}]=0$$
A: In case $[x]$ is intended to be an integer part of $x$, you have
$$
  \lim_{x\to 0}x\left[\frac1x\right] = \lim_{y\to\infty}\frac{[y]}{y} = \lim_{y\to\infty}\frac{y -\{y\}}y = 1.
$$
