Rozwiązanie:
[tex]a)\\[/tex]
[tex]\lim_{n \to \infty} \frac{n^{2}-1}{n^{3}+3n+2} = \lim_{n \to \infty} \frac{n^{3}(\frac{1}{n}-\frac{1}{n^{3}} )}{n^{3}(1+\frac{3}{n^{2}}+\frac{2}{n^{3})} } =\frac{0}{1}=0[/tex]
[tex]b)[/tex]
[tex]\lim_{n \to \infty} \frac{\sqrt{n^{2}-1} }{n}= \lim_{n \to \infty} \frac{\sqrt{n^{2}(1-\frac{1}{n^{2}}) } }{n} = \lim_{n \to \infty} \frac{n\sqrt{(1-\frac{1}{n^{2}})} }{n}=1[/tex]
[tex]c)[/tex]
[tex]\lim_{n \to \infty} (1+\frac{2}{n})^{n}=e^{2}[/tex]
