Evaluate $\lim_{x \to 0} x^2\left(1+2+3+...+\left[ \frac{1}{|x|} \right] \right)$ I was thinking about the squeeze theorem here. We can denote the $\left[\frac{1}{|x|}\right] =n$, and then try something like:
$$x^2(1+2+3+...+(n-1)+(n-1)) \leq x^2 \frac{n(n+1)}{2} \leq x^2(1+2+3+...+(n-1)+(n+1))$$
But I don't know what to do with $x^2$. Given that $\left[ \frac{1}{|x|} \right]=n$, how do we proceed to find $x^2$?

Hint
First we may suppose $x> 0$ as well. Next, 
$$\biggl\lfloor\frac1x\biggr\rfloor=n\iff n\le\frac1x<n+1, \enspace\text{ so }\enspace x\dots$$
