Oblicz granice:[/latex][latex] lim_{n o infty} ( frac{1+2+3+...+n}{n-2 n^{2} } )= \ lim_{n o infty} ( frac{2 n^{4}- n^{2}+2 }{3 n^{3}+6 } )=[/latex]

Oblicz granice:[/latex][latex] lim_{n o infty} ( frac{1+2+3+...+n}{n-2 n^{2} } )= \ lim_{n o infty} ( frac{2 n^{4}- n^{2}+2 }{3 n^{3}+6 } )=[/latex]
Odpowiedź

w 1 na górze jest ciąg arytmetyczny [latex]S_n=frac{a_1+a_n}{2} cdot n = frac{1+n}{2} cdot n = frac{n+n^2}{2}[/latex] [latex]frac{frac{n+n^2}{2}}{n-2n^2}=frac{n+n^2}{2n-4n^2}=frac{n^2(frac{1}{n}+1)}{n^2(frac{2}{n}-4)}=frac{frac{1}{n}+1}{frac{2}{n}-4}[/latex] [latex] lim_{n o infty} (frac{frac{1}{n}+1}{frac{2}{n}-4}) = frac{0+1}{0-4}=-frac{1}{4} [/latex] 2. [latex]frac{2n^4-n^2+2}{3n^3+6}=frac{n^3(2n-frac{1}{n}+frac{2}{n^3})}{n^3(3+frac{6}{n^3})}=frac{2n-frac{1}{n}+frac{2}{n^3}}{3+frac{6}{n^3}}[/latex] [latex] lim_{n o infty} (frac{2n-frac{1}{n}+frac{2}{n^3}}{3+frac{6}{n^3}})=frac{infty-0+0}{3+0}=frac{infty}{3}=infty[/latex]

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