Odpowiedź:
1°
[tex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{4-3n}{2n+3} = \lim_{n \to \infty} \frac{n(\frac{4}{n}-3) }{n(2+\frac{3}{n}) } =\frac{0-3}{2+0} =-\frac{3}{2}[/tex]
2°
[tex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{5n^2+3n-1}{2n^2-3n+2} = \lim_{n \to \infty} \frac{n^2(5+\frac{3}{n}-\frac{1}{n^2}) }{n^2(2-\frac{3}{n}+\frac{2}{n^2} ) } =\frac{5+0-0}{2-0+0} =\frac{5}{2}[/tex]
3°
[tex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3n^3+2n+1}{2n^2-3} = \lim_{n \to \infty} \frac{n^3(3+\frac{2}{n^2}+\frac{1}{n^3}) }{n^2(2-\frac{3}{n^2}) } = \lim_{n \to \infty} \frac{n(3+\frac{2}{n^2}+\frac{1}{n^3} ) }{2-\frac{3}{n^2} } = \infty[/tex]
4°
[tex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2+2n-1}{3n+2} = \lim_{n \to \infty} \frac{n^2(1+\frac{2}{n}-\frac{1}{n^2}) }{n(3+\frac{2}{n}) } = \lim_{n \to \infty} \frac{n(1+\frac{2}{n}-\frac{1}{n^2}) }{3+\frac{2}{n} } = \infty[/tex]
