What is the valuation of $\lim_{x\to0}\ x^2\left(1+2+3+…+\left[\frac1{|x|}\right]\right)$?

Compute the following limitation: $\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$.

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What is $[...]$? Is it the floor function? –  Babak S. Dec 29 '12 at 8:56

Recall that $$1 + 2 + 3 + \cdots + \cdots + n = \dfrac{n(n+1)}2$$ If $\vert x \vert \in \left(\dfrac1{n+1}, \dfrac1n\right]$, we have $\lfloor 1/\vert x \vert \rfloor = n$ and hence we get that $$1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor = \dfrac{\lfloor 1/\vert x \vert \rfloor(\lfloor 1/\vert x \vert \rfloor+1)}2 = \dfrac{n(n+1)}2$$ Then $$x^2 \left(1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor \right) = \dfrac{x^2n(n+1)}2 \in \left(\dfrac{n}{2(n+1)}, \dfrac{n+1}{2n} \right]$$ Hence, as $n \to \infty$ i.e. as $\vert x \vert \to 0$, we get the limit as $1/2$.
Very nice. Now it seems obvious that one wants to relate the dots in the expression to $x$ via $|x| \in (\frac{1}{n+1}, \frac{1}{n}]$. –  Rudy the Reindeer Dec 29 '12 at 9:05