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 $x$.

$$1>\frac{i}{x}-\left[\frac{i}{x}\right]\ge0$$
For $x>0$, multiply by $x$
$$x>i-x\left[\frac{i}{x}\right]\ge0$$
Sum for $1\le i\le10$
$$10x>55-x\sum \left[\frac{i}{x}\right]\ge0$$
Let $x$ tend to $0$, then $x\sum \left[\frac{i}{x}\right]$ tends to $55$.
