# Thread: Limit of a floor function

1. ## Limit of a floor function

$\displaystyle \lim_ {x \rightarrow \00} (x[1/x])=?$
Hello,
the squared brackets mean floor function.

Thanks,
Michael

2. We have $\displaystyle \displaystyle \lim_{x\to 0 }x\lfloor \frac 1x\rfloor =\lim_{t\to\infty}\frac 1t\lfloor t\rfloor$. Now you can use a well-known inequality.

3. To make notation easier say $\displaystyle f(x) = x\left\lfloor {\frac{1}{x}} \right\rfloor$.
Look at the limit $\displaystyle \displaystyle\lim _{x \to 0^ + } f(x)$.
For each $\displaystyle n\in\mathbb{Z}^+$ if $\displaystyle x\in\left(\frac{1}{n+1},\frac{1}{n}\right)$ then $\displaystyle n<\frac{1}{x}<n+1$.
So it follows that $\displaystyle \left\lfloor {\frac{1}{x}} \right\rfloor={n}$ which implies that $\displaystyle \frac{n}{{n + 1}} < f(x) < 1$

So what is the limit on the right?

You can do the limit on the left.

4. Thank a lot, but I cannot understand what are you trying to say, can you please simplify your answer?
I remind you the the squared brackets mean floor(1/X)

5. Originally Posted by msokol89
Thank a lot, but I cannot understand what are you trying to say, can you please simplify your answer?
I remind you the the squared brackets mean floor(1/X)
Do you know anything about the floor function?
How does it work?

BTW, $\displaystyle \lfloor x\rfloor$ is the standard notation.

6. Yes, it gives the highest integer that is lower than a specific number,
My problem is that I am not that familiar with the official notations.

7. Originally Posted by msokol89
Yes, it gives the highest integer that is lower than a specific number, My problem is that I am not that familiar with the official notations.
Here is a web reference.