Rozwiązanie:
[tex]$ \lim_{n \to \infty} \Big[\frac{1}{n} {n \choose 1}+\frac{1}{n^{2}} {n \choose 2} +\frac{1}{n^{3} } {n \choose 3} \Big]=[/tex]
[tex]$= \lim_{n \to \infty} \Big[\frac{1}{n} \cdot \frac{n!}{1!(n-1)!} +\frac{1}{n^{2}} \cdot \frac{n!}{2!(n-2)!} + \frac{1}{n^{3}} \cdot \frac{n!}{3!(n-3)!} \Big]=[/tex]
[tex]$= \lim_{n \to \infty} \Big[\frac{1}{n} \cdot n+\frac{1}{n^{2}} \cdot \frac{n(n-1)}{2} +\frac{1}{n^{3}} \cdot \frac{n(n-1)(n-2)}{6} \Big]=[/tex]
[tex]$= \lim_{n \to \infty} \Big[1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{6n^{2}} \Big]=[/tex]
[tex]$= \lim_{n \to \infty} 1+ \lim_{n \to \infty} \frac{n\Big(1-\frac{1}{n} \Big)}{2n} + \lim_{n \to \infty} \frac{n^{2}\Big(1-\frac{1}{n} \Big)\Big(1-\frac{2}{n} \Big)}{6n^{2}} =[/tex]
[tex]$=1+\frac{1}{2} +\frac{1}{6} =\frac{5}{3}[/tex]
