[tex] \lim_{n \to \infty} \frac{1+2+3+...+n}{2n^2+3}}= \lim_{n \to \infty} \frac{ \frac{n(n+1)}{2} }{2n^2+3}[/tex], simplificand fortat cu n², se obtine : =[tex] \lim_{n \to \infty} \frac{1+2+3+...+n}{2n^2+3}}= \lim_{n \to \infty} \frac{ \frac{n(n+1)}{2} }{2n^2+3}= \lim_{n \to \infty} \frac{1+ \frac{1}{n} }{4+ \frac{6}{n^2}} = \frac{1}{4}[/tex]
