Matematyka.pl

oblicz granice

1. \(\displaystyle{ \lim_{n \to \infty}}\)\(\displaystyle{ \left( \sqrt{2+n+n ^{2}- \sqrt{\left( n+1\right) ^{2}+2 } } \right)}\)

2. \(\displaystyle{ x_n= \left( \frac{4n+2}{4n+3} \right) ^{5-8n}}\)

3.\(\displaystyle{ a_n= \left( \frac{n}{n+1} \right) ^{ \frac{n+2}{2} }}\)

zizu_56 pisze:
3.\(\displaystyle{ \lim_{ n\to \infty} = \left( \frac{n}{n+1} \right) ^{ \frac{n+2}{2} }}\)

\(\displaystyle{ ...= \lim_{x \to \infty} [(1+ \frac{-1}{n+1}) ^{ \frac{n+1}{-1} }] ^{ \frac{-n-2}{2n+2} }=e ^{- \frac{1}{2} }}\)-- 10 lutego 2009, 22:58 --

zizu_56 pisze:
2. \(\displaystyle{ xn= \left( \frac{4n+2}{4n+3} \right) ^{5-8n}}\)

\(\displaystyle{ ...= \lim_{ n\to \infty} [(1+ \frac{-1}{4n+3}) ^{ \frac{4n+3}{-1} }] ^{ \frac{8n-5}{4n+3} } =e ^{2}}\)

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

Matematyka.pl

obliczyć granice

obliczyć granice

obliczyć granice

obliczyć granice