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Oblicz granice:

: 11 lis 2007, o 11:13
autor: FK
kożystając z def. liczby e oraz z tw. o granicy podciągu oblicz granice:

\(\displaystyle{ \lim_{n\to\infty}\left(\frac{5n+2}{5n+1}\right) ^{15n}}\)

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

\(\displaystyle{ \lim_{n\to\infty}\left(\frac{3n+1}{3n+2}\right) ^{6n}}\)

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

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

Oblicz granice:

: 11 lis 2007, o 11:25
autor: soku11
1.
\(\displaystyle{ \lim_{n\to\infty}\left(\frac{5n+1+1}{5n+1}\right) ^{15n} =
\lim_{n\to\infty}\left(1+\frac{1}{5n+1}\right) ^{15n} =
\lim_{n\to\infty} ft[\left(1+\frac{1}{5n+1}\right) ^{5n+1}\right]^{\frac{15n}{5n+1}} =e^3}\)


2.
\(\displaystyle{ \lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right) ^{n} =
\lim_{n\to\infty} ft[\left(1+\frac{-1}{n+1}\right) ^{-(n+1)}\right]^{\frac{-n}{n+1}} =e^{-1}}\)


3.
\(\displaystyle{ \lim_{n\to\infty}\left(\frac{3n+2-1}{3n+2}\right) ^{6n} =
\lim_{n\to\infty}\left(1+\frac{-1}{3n+2}\right) ^{6n} =
\lim_{n\to\infty} ft[\left(1+\frac{-1}{3n+2}\right) ^{-(3n+2)}\right]^{\frac{-6n}{3n+2}} =
e^{-2}}\)



4.
\(\displaystyle{ \lim_{n\to\infty}\left(\frac{3n+1-1}{3n+1}\right) ^{n} =
\lim_{n\to\infty}\left(1+\frac{-1}{3n+1}\right) ^{n} =
\lim_{n\to\infty}\left[\left(1+\frac{-1}{3n+1}\right) ^{-(3n+1)}\right]^{\frac{-n}{3n+1}} =
e^{-\frac{1}{3}}}\)



POZDRO