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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

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

  • 1
    $[x]=0$ 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.2011-12-03

2 Answers 2

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Your solution works with the correction I noted in the comments:

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. } $