Odpowiedź:
a)
[tex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3+6+9+...+3n}{5n^2-2}= \lim_{n \to \infty} \frac{n^2(\frac{3}{n^2}+\frac{6}{n^2}+\frac{9}{n^2}+...+\frac{3}{n} )}{n^2(5-\frac{2}{n^2}) }=\\= \lim_{n \to \infty} \frac{\frac{3}{n^2}+\frac{6}{n^2}+\frac{9}{n^2}+...+\frac{3}{n}}{5-\frac{2}{n^2}}=\frac{1}{5}[/tex]
poniewaz:
[tex]\lim_{n \to \infty} \frac{3}{n^2} =0\\\\ \lim_{n \to \infty} \frac{6}{n^2} =0\\\\ \lim_{n \to \infty} \frac{9}{n^2} =0\\\\ \lim_{n \to \infty} \frac{3}{n} =0\\\\ \lim_{n \to \infty} \frac{2}{n^2} =0\\[/tex]
b)
[tex]\lim_{n \to \infty} \frac{4^n-5^n}{4^{n+1}+4*5^n}= \lim_{n \to \infty} \frac{4^n-5^n}{4(4^{n}+5^n)}= \lim_{n \to \infty} \frac{1}{4} \frac{4^n-5^n}{4^{n}+5^n}=\\[/tex]
[tex]=\frac{1}{4} \lim_{n \to \infty} \frac{4^n-5^n}{4^{n}+5^n}*\frac{\frac{1}{5^n} }{\frac{1}{5^n} }=\frac{1}{4} \lim_{n \to \infty} \frac{4^n*\frac{1}{5^n} -5^n*\frac{1}{5^n} }{4^{n}*\frac{1}{5^n}+5^n*\frac{1}{5^n}}=\frac{1}{4} \lim_{n \to \infty} \frac{(\frac{4}{5})^n-1 }{\frac{4}{5})^n+1}=\\=\frac{1}{4} *(-1)=-\frac{1}{4}[/tex]
poniewaz
[tex]\lim_{n \to \infty} (\frac{4}{5})^n=0[/tex]
