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rymoholiko · Post autor: **rymoholiko** » 5 lut 2009, o 19:21

\(\displaystyle{ \lim_{ n\to \infty } \left( \frac{3n+1}{3n+2} \right)^{5n}}\)

abc666 · Post autor: **abc666** » 5 lut 2009, o 21:24

\(\displaystyle{ \lim_{ n\to \infty } \left( \frac{3n+1}{3n+2} \right)^{5n}= \left(1+ \frac{-1}{3n+2} \right) ^{5n}=\left(1+ \frac{-1}{3n+2} \right) ^{(3n+2) \frac{5n}{3n+2} }=e^{- \frac{5}{3} }}\)

meninio · Post autor: **meninio** » 5 lut 2009, o 21:25

\(\displaystyle{ \lim_{ n\to \infty } \left( \frac{3n+1}{3n+2} \right)^{5n}=\lim_{ n\to \infty } \left( \frac{3n+2-1}{3n+2} \right)^{5n}=\lim_{ n\to \infty } \left(1+ \frac{-1}{3n+2} \right)^{5n}= \\ \\ = \lim_{ n\to \infty } \left(1+ \frac{-1}{3n+2} \right)^{\frac{3n+2}{-1} \cdot \frac{-1}{3n+2} \cdot 5n}=\lim_{ n\to \infty } e^{ \frac{-5n}{3n+2}}=e^{-\frac{5}{3}}}\)