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 $x$.
 A: 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).
A: Generally:
$$\lim_{x \to 0^+} x \left[ \dfrac Cx \right] = \lim_{x \to 0^+} x\left( \frac Cx - \left\{\frac C x\right\}\right) = \lim_{x \to 0^+} \left(C-x \left\{\frac px\right\}\right) = C$$ 
Take a look here too: How can calculate this limit?
A: Let $$\dfrac1x=I+f$$ where $0\le f<1$ and $I$ is an integer
$$x\sum_{r=1}\left[\dfrac rx\right]=\dfrac{n(n+1)}2\cdot\dfrac I{I+f}$$
Now as $x\to0,I\to+\infty$
A: $$1>\frac{i}{x}-\left[\frac{i}{x}\right]\ge0$$
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 $55$.
A: We can convert to
$$\lim_{t\to\pm\infty}\frac{\lfloor t\rfloor+\lfloor2t\rfloor+\lfloor3t\rfloor+\cdots\lfloor10t\rfloor}t=\\
\lim_{t\to\pm\infty}\left(1+2+3+\cdots10-\frac{\{t\}+\{2t\}+\{3t\}+\cdots\{10t\}}t\right).$$
The second term vanishes because the numerator remains in range $[0,10)$.

Notation: $t=\lfloor t\rfloor+\{t\}$.
A: From $x-1<[x]\leq x$, we get
$$
x\left(\sum^{10}_{k=1}\frac{k}{x}-10\right)<x\sum^{10}_{k=1}\left[\frac{k}{x}\right]\leq x\sum^{10}_{k=1}\frac{k}{x}
$$
Taking the limit $x\rightarrow 0$, we find easily that 
$$
\lim_{x\rightarrow 0}x\sum^{10}_{k=1}\left[\frac{k}{x}\right]=\sum^{10}_{k=1}k=\frac{10(10+1)}{2}=55
$$ 
