Question

Oblicz:
[tex]a) \lim_{n \to \infty} (3-4n^2-2n^3)\\\\b) \lim_{n \to \infty} \frac{n(n+1)(n-4)}{4n+n^3}\\\\c) \lim_{n \to \infty} \frac{6+12+18+...+6n}{2n^2-1}\\[/tex]

Answer 1

[tex]a)\lim_{n \to \infty} (3-4n^{2}-2n^{3})=3-\infty-\infty=-\infty\\[/tex]

[tex]b)\lim_{n \to \infty} \frac{n(n+1)(n-4)}{4n+n^{3}} =\lim_{n \to \infty} \frac{n(n^{2}-3n-4)}{4n+n^{3}}=\lim_{n \to \infty} \frac{n^{3}-3n^{2}-4n}{4n+n^{3}}=\lim_{n \to \infty} \frac{n^{3}(1-\frac{3}{n}-\frac{4}{n^{2}}) }{n^{3}(1+\frac{4}{n}) }=\frac{1}{1} =1[/tex]

[tex]c) \lim_{n \to \infty}\frac{6+12+18+...+6n}{2n^{2}-1} \\\\a_{1}=6\\r=6\\a_{n}=6+6(n-1)=6+6n-6=6n\\S_{n}=\frac{6+6n}{2} *n=(3+3n)m=3n^{2}+3n\\\\\lim_{n \to \infty}\frac{6+12+18+...+6n}{2n^{2}-1} =\lim_{n \to \infty}\frac{3n^{2}+3n}{2n^{2}-1} = \lim_{n \to \infty}\frac{n^{2}(3+\frac{3}{n}) }{n^{2}(2-\frac{1}{n^{2}}) } =\frac{3}{2}[/tex]