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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

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

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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 $\frac 1 2$

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