Odpowiedź:
a) 1
b) 0
c) 0
d) 0
Szczegółowe wyjaśnienie:
[tex]a) \lim_{n \to \infty} \frac{2n-1}{2n}= \lim_{n \to \infty} \frac{2n(1-\frac{1}{2n}) }{2n} = \lim_{n \to \infty} 1-\frac{1}{2n} =1\\b) \lim_{n \to \infty} \frac{3n-2}{(3n-1)(2n+1)}= \lim_{n \to \infty} \frac{n(3-\frac{2}{n}) }{n(3-\frac{1}{n})*n(2+\frac{1}{n}) } = \lim_{n \to \infty} \frac{3-\frac{2}{n} }{n(3-\frac{1}{n})(2+\frac{1}{n} ) } =0\\c) \lim_{n \to \infty} \frac{n}{n^{2} } = \lim_{n \to \infty} \frac{n^{2}(\frac{1}{n}) }{n^{2}} = \lim_{n \to \infty} \frac{1}{n} =0\\\\[/tex]
[tex]d) \lim_{n \to \infty} \frac{n^{2} }{1-3n^{3} } = \lim_{n \to \infty} \frac{n^{3}(\frac{1}{n} ) }{n^{3}(\frac{1}{n^{3} }-3 ) } = \lim_{n \to \infty} \frac{\frac{1}{n}}{ \frac{1}{n^{3} }-3 } } =0[/tex]
