Matematyka
FowlPS
2017-06-26 01:52:19
Kto pomoze? Obliczyć granice ciagu lim [(2n-5)/(2n+3)]^7n
Odpowiedź
MarzenaMazi
2017-06-26 03:37:42

[latex]limlimits_{n o infty} (frac{2n-5}{2n+3})^{7n} = limlimits_{n o infty} (frac{2n+3}{2n+3} + frac{-8}{2n+3} )^{7n} = limlimits_{n o infty} (1+ frac{-8}{2n+3})^{7n} =[/latex] [latex]limlimits_{n o infty} (1+ frac{-8}{2n+3})^{frac{7}{2}(2n+3)-frac{21}{2}} = limlimits_{n o infty} [ (1+frac{-8}{2n+3})^{frac{7}{2}(2n+3)} cdot (1+frac{-8}{2n+3})^{-frac{21}{2}})=[/latex] [latex]limlimits_{n o infty} [ [ (1+frac{-8}{2n+3})^{2n+3}]^frac{7}{2} cdot(1+frac{-8}{2n+3})^{-frac{21}{2}}]=(e^{-8})^frac{7}{2} cdot 1^{-frac{21}{2}} = e^{-28} cdot 1=[/latex] [latex]e^{-28}[/latex] 2: [latex]limlimits_{n o infty} (frac{2n+3}{2n-5})^{7n} = limlimits_{n o infty} (frac{2n-5}{2n-5} + frac{8}{2n-5} )^{7n} = limlimits_{n o infty} (1+ frac{8}{2n-5})^{7n} =[/latex] [latex]limlimits_{n o infty} (1+ frac{8}{2n-5})^{frac{7}{2}(2n-5)+frac{35}{2}} = limlimits_{n o infty} [ (1+frac{8}{2n-5})^{frac{7}{2}(2n-5)} cdot (1+frac{8}{2n-5})^frac{35}{2})=[/latex] [latex]limlimits_{n o infty} [ [ (1+frac{8}{2n-5})^{2n-5}]^frac{7}{2} cdot(1+frac{8}{2n-5})^frac{35}{2}]=(e^{8})^frac{7}{2} cdot 1^frac{35}{2} = e^{28} cdot 1=[/latex] [latex]e^{28}[/latex]

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