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