Find the limit $\lim_{x\to 0} x\left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right)$

Question

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 $x$.

Answer 1

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).

Answer 2

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?

Answer 3

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 $x\to0,I\to+\infty$

Answer 4

$$1>\frac{i}{x}-\left[\frac{i}{x}\right]\ge0$$ 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 $55$.

Answer 5

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 $[0,10)$.

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Notation: $t=\lfloor t\rfloor+\{t\}$.

Answer 6

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 $$ \lim_{x\rightarrow 0}x\sum^{10}_{k=1}\left[\frac{k}{x}\right]=\sum^{10}_{k=1}k=\frac{10(10+1)}{2}=55 $$