# Finding limit involving floor function.

Finding $\displaystyle \lim_{x\rightarrow 0}x^2\bigg(1+2+3+\cdots \cdots +\bigg\lfloor \frac{1}{|x|}\bigg\rfloor \bigg)$, where $\lfloor x \rfloor$ is a floor function of $x$

Attempt: put $\displaystyle x = \frac{1}{y}$

so $\displaystyle \lim_{y\rightarrow \infty}\frac{1+2+3+\cdots \cdots \lfloor y \rfloor }{y^2}$

could some help me

• The sum is not clear at all, what is $x$? How the sum is defined? How many elements have the sum? – Masacroso Feb 25 '17 at 8:41
• You know what $1+2+\cdots+[y]$ is, and you know $y-1<[y]\le y$, so put these together. – Gerry Myerson Feb 25 '17 at 9:11

The sum rewrites $1+2+ \dots +\lfloor y\rfloor = \frac{\lfloor y\rfloor (\lfloor y\rfloor +1)}{2}$. Therefore, the studied function is asymptotically equivalent to $\frac{1}{2} \left(\frac{\lfloor y\rfloor}{y}\right)^2$, which tends towards $1/2$ at infinity.
We know that $y - 1 \lt \lfloor y \lfloor \le y$. From here, $\frac {(y-1)y} 2 \lt \frac {\lfloor y\rfloor (\lfloor y\rfloor +1)} {2} \le \frac {y(y + 1)} 2$ therefore, using the squeezing theorem, the limit is $\frac 1 2$