Oblicz granicę ciągu
\(\displaystyle{ \lim_{n \to \infty } \frac{\log _{3}[n(n-6) ^{5}] }{\log _{81}n ^{6} }}\)
Doszlam do czegos takiego
\(\displaystyle{ \lim_{n \to \infty } \frac{\log _{3}[n(n-6) ^{5}] }{ \frac{1}{4} \log _{3}n ^{6} }}\)
i nie wiem co dalej
granica ciągu
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Quester
- Użytkownik
- Posty: 25
- Rejestracja: 18 cze 2011, o 02:41
- Płeć: Mężczyzna
- Lokalizacja: Polska
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- Pomógł: 2 razy
granica ciągu
\(\displaystyle{ \lim_{n \to \infty } \frac{\log _{3}[n(n-6) ^{5}] }{ \frac{1}{4} \log _{3}n ^{6} } = 4\lim_{n \to \infty } \log _{n^6}[n(n-6) ^{5}]}\)
\(\displaystyle{ \log _{n^6}[n(n-6) ^{5}] < \log _{n^6}n ^{6} = 1}\)
\(\displaystyle{ \log _{n^6}[n(n-6) ^{5}] > \log _{n^6}[n \left( \frac{n}{2} \right) ^{5}] \hbox{ dla } n > 12}\)
\(\displaystyle{ \lim_{n \to \infty } \log _{n^6}[n\left( \frac{n}{2} \right) ^{5}] = \lim_{n \to \infty } \log _{n^6} \left( \frac{n ^{6}}{32}\right)= \lim_{n \to \infty } \log _{n^6}n ^{6} - \lim_{n \to \infty } \log _{n^6} 32 = 1}\)
\(\displaystyle{ \log _{n^6}[n \left( \frac{n}{2} \right) ^{5}] < \log _{n^6}[n(n-6) ^{5}] < 1 \hbox{ dla } n > 12}\)
\(\displaystyle{ \lim_{n \to \infty }\log _{n^6}[n(n-6) ^{5}] = 1}\)
\(\displaystyle{ \log _{n^6}[n(n-6) ^{5}] < \log _{n^6}n ^{6} = 1}\)
\(\displaystyle{ \log _{n^6}[n(n-6) ^{5}] > \log _{n^6}[n \left( \frac{n}{2} \right) ^{5}] \hbox{ dla } n > 12}\)
\(\displaystyle{ \lim_{n \to \infty } \log _{n^6}[n\left( \frac{n}{2} \right) ^{5}] = \lim_{n \to \infty } \log _{n^6} \left( \frac{n ^{6}}{32}\right)= \lim_{n \to \infty } \log _{n^6}n ^{6} - \lim_{n \to \infty } \log _{n^6} 32 = 1}\)
\(\displaystyle{ \log _{n^6}[n \left( \frac{n}{2} \right) ^{5}] < \log _{n^6}[n(n-6) ^{5}] < 1 \hbox{ dla } n > 12}\)
\(\displaystyle{ \lim_{n \to \infty }\log _{n^6}[n(n-6) ^{5}] = 1}\)