# Convergence of limit with integer part of $x$

Good day! I tried to solve this problem;the process is correct?

The problem si:

Let $x\in\mathbb{R}$. With $[x]$ denote the integer part of $x$. Calculate

$$\lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr),\qquad k\in \mathbb{N}.$$

My solution:

$$\lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr)=\lim_{x\to 0^+} \ x^2\Biggr(\frac{1}{[x]}+\frac{2}{[x]}+\dots + \frac{k}{[x]}\Biggr)=$$

$$=\lim_{x\to 0^+} \ x^2\Biggr(\frac{1+2+3+\dots+k}{[x]}\Biggr)=\lim_{x\to 0^+} \ x^2 \ \sum_{j=1}^{k}\frac{j}{[x]}=\lim_{x\to 0^+} \frac{x^2 }{[x]} \ \sum_{j=1}^{k} \ {j}$$

Now we know that: $$x-1<[x]\le x.$$ so $$\frac{1}{x}\le\frac{1 }{[x]}\le\frac{1}{x-1}$$ because ${x\to 0^+}$. So $$\frac{x^2}{x}\le\frac{x^2 }{[x]}\le\frac{x^2}{x-1}$$ passing to the limit, and applying the comparison, we have

$$\lim_{x\to 0^+} \frac{x^2 }{[x]} \ \sum_{j=1}^{k} \ {j}=0$$

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$[x]=0$ if $0\le x<1$; so, your first step is not valid. But you can replace $[1/x]$ with just $1/x$ at the expense of introducing an inequality. –  David Mitra Dec 3 '11 at 9:08
I note your acceptance rate is zero. Do you know about accepting answers on this website? Maybe you should look into it. It helps the website work better. –  Gerry Myerson Dec 3 '11 at 9:17
Since $[y]\le y$ for $y\ge0$, \eqalign{0&\le \lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr)\cr &\le \lim_{x\to 0^+} \Biggr(x^2 ( \frac{1}{x} + \frac{2}{x} +\dots + \frac{k}{x} )\Biggr)\cr &= \lim_{x\to 0^+} [ x (1+2+3+\cdots+ k)]\cr &=0. }