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.

In case $[x]$ is intended to be an integer part of $x$, you have $$ \lim_{x\to 0}x\left[\frac1x\right] = \lim_{y\to\infty}\frac{[y]}{y} = \lim_{y\to\infty}\frac{y -\{y\}}y = 1. $$

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$

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 $$\lim_{x \to 0} x \left\lfloor \dfrac1x \right\rfloor = 1$$