Find the limit $\lim_{x\to 0} x\left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right)$ Can someone help me finding the following limit 
$$ \lim_{x\to 0} x\left(\left\lfloor\frac{1}{x}\right\rfloor +\left\lfloor\frac{2}{x}\right\rfloor +\cdots \left\lfloor\frac{10}{x}\right\rfloor\right)$$
I can somehow guess the limit will be $55$, as $\lim_{x\to 0}x\left\lfloor\frac{1}{x}\right\rfloor=1$. But, I am not able to prove it.
Note: $\left\lfloor x\right\rfloor$ denotes the greatest integer less than or equal to $x$.

Hint Since $u-1 < \lfloor u \rfloor \leq u$, you have
$$ \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)-10 \leq \left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right) \leq \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)$$
Therefore,
$$ \frac{55}{x} -10 \leq \left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right) \leq \frac{55}{x}$$
Now multiply both sides by $x$, splitting the problem into $x >0$ and $x <0$ (since in the second case the inequality flips when you multiply).
