<s>[INST] Limit of sequence with greatest integer function Calculate $\lim \limits_{x \to 0} \ x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big)$, where [.] represents greatest integer function.
Since we know that $x-1\lt [x]\leq x,$ I tried squeezing the sequence between $\ x^2\Big(1+2+3+....+\ \frac{1}{\vert x \vert} -1\Big)$ and $\ x^2\Big(1+2+3+....+\ \frac{1}{\vert x \vert}\Big)$ which gives me the limit as $0$. But the correct answer is given as $\frac{1}{2}$ .
 [/INST] The sum of the first $n$ integers is $\frac {n(n+1)}2$. Thus, with $n=\Big[\frac{1}{\vert x \vert}\Big]$,
$$x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big) = x^2\frac {n(n+1)}2$$
Since, by definition of $n$, we have $n\leq \frac 1 x < n+1$
$$ x^2\frac{\left(\frac 1 x - 1\right) \frac 1 x}2< x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big)  \leq x^2\frac{\frac 1 x \left(\frac 1 x +1\right)}2$$
Taking the limit (squeezing) gives that the limit is $\frac 1 2$.
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