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ona123

Rejestracja: 18 Sep 2009
Offline Ostatnio: May 28 2018 15:53
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Moje tematy

Granica ciągu

02.11.2011 - 22:11

\frac{2}{\sqrt{4n^2+9n}-2n}=\frac{2}{\frac{4n^2+9n-4n^2}{\sqrt{4n^2+9n}+2n}}=\frac{2}{\frac{9n}{\sqrt{4n^2+9n}+2n}}=\frac{2}{\frac{9n}{\sqrt{n^2(4+\frac{9}{n})}+2n}}=\frac{2}{\frac{9n}{\sqrt{n^2*\sqrt{4+\frac{9}{n}}}+2n}}=\frac{2}{\frac{9n}{n\sqrt{4+\frac{9}{n}}+2n}}=\frac{2}{\frac{9}{\sqrt{4+\frac{9}{n}}+2}}=\frac{2}{\frac{9}{4}}=\frac{8}{9}

Granica ciągu

02.11.2011 - 21:19

Proszę o sprawdzenie.

a. \sqrt{n^2+4n}-n= \frac{n^2+4n-n^2}{\sqrt{n^2+4n}+n=\frac{4n}{n^2(\sqrt{1+\frac{4}{n})+n}}=\frac{4}{\sqrt{1+\frac{4}{n}+1=\frac{4}{2}=2

Granica ciągu

31.10.2011 - 23:37

Proszę o sprawdzenie przykładów.

a) \frac{(3n+1)^2}{[(1-n)(1+n)]^2}=\frac{(3n+1)(3n+1)}{1-n^2}=\frac{3n+1}{1-n^2}=\frac{3n(1+\frac{1}{3n)}}{n^2(\frac{1}{n^2}-1)}=\frac{3n+\frac{1}{3n}}{n^2}= \frac{3n}{n^2}=\frac{3}{n}=0



b)\frac{2}{\sqrt{4n^2+9n}-2n}=\frac{2}{\frac{4n^2+9n-2n^2}{4n^2+9n+2n}=\frac{2}{-n}=0


c) \frac{3^n^-^2-2*5^2^n^-^1+3}{4^n^+^3+5^n^+^4-7}= \frac{3^n*3^-^2-2*25^n+3}{4^n*64+5^n*625-7}=\frac{3^n*\frac{1}{9}-2*25^n+3}{4^n*64+5^n*625-7}=\frac{25^n((\frac{3}{25})^n*\frac{1}{9}-2+\frac{3}{25^n})}{5^n((\frac{3}{5})^n*64+625-\frac{7}{5^n})}=\frac{-2}{625}*(\frac{25}{5})^n=0