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granica ciągu
: 26 sty 2008, o 22:32
autor: czbk
\(\displaystyle{ \lim_{n\to\infty} ft(\frac{2n+7}{n+3}\right)^{(3n+1)/n}}\)
Przepraszam, moj blad w latexie P.S dzieki za rozwiazanie tego, czy moglbyc zrobic i to ?
granica ciągu
: 26 sty 2008, o 22:38
autor: scyth
\(\displaystyle{ 2^{3n} = ft( \frac{2n+6}{n+3} \right)^{3n} < ft( \frac{2n+7}{n+3} \right)^{3n+\frac{1}{n}} \\
\lim_{n \to } 2^{3n} = \\
\lim_{n \to } ft( \frac{2n+7}{n+3} \right)^{3n+\frac{1}{n}} = }\)
no to teraz poprawione:
\(\displaystyle{ \left( \frac{2n+7}{n+3} \right)^{\frac{3n+1}{n}} = ft(2+\frac{1}{n+3}\right)^3 ft(2+\frac{1}{n+3}\right)^{\frac{1}{n}}\\
\lim_{n \to } ft( \frac{2n+7}{n+3} \right)^{\frac{3n+1}{n}} =
\lim_{n \to } ft(2+\frac{1}{n+3}\right)^3 ft(2+\frac{1}{n+3}\right)^{\frac{1}{n}} = 2^3 1 = 8}\)