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Granica ciągów

: 14 mar 2018, o 14:05
autor: Wojtek Jerzy
Witam, mam problem z 2 zadaniami z granicy ciągów.
zad. 1:
\(\displaystyle{ \lim_{ n \to \infty } \frac{\sqrt{{2n ^{2}-1 }}+{ \sqrt{n ^{2}+3n+3 }}}{ \sqrt[3]{2n ^{3}-1 } }}\)

Zad.2:
\(\displaystyle{ \lim_{n \to \infty } \frac{ {n \choose 2} }{n ^{2} +3n -1}}\)

Dziękuje za pomoc

Re: Granica ciągów

: 14 mar 2018, o 14:37
autor: Bierut
Ad. 1
\(\displaystyle{ \lim_{ n \to \infty } \frac{\sqrt{{2n ^{2}-1 }}+{ \sqrt{n ^{2}+3n+3 }}}{ \sqrt[3]{2n ^{3}-1 } }=
\lim_{n\to\infty} \frac{\sqrt{n^{2}\left(2-\frac{1}{n^2}\right)}+\sqrt{n^{2}\left(1+\frac{3}{n}+\frac{3}{n^2}\right)}}{\sqrt[3]{n^{3}\left(2-\frac{1}{n^3}\right)}}= \\ =
\lim_{n\to\infty} \frac{n\left(\sqrt{2-\frac{1}{n^2}}+\sqrt{1+\frac{3}{n}+\frac{3}{n^2}}\right)}{n\sqrt[3]{2-\frac{1}{n^3}}}=
\frac{\sqrt{2}+1}{\sqrt[3]{2}}}\)


Ad. 2
\(\displaystyle{ \lim_{n \to \infty } \frac{ {n \choose 2} }{n ^{2} +3n -1}=
\lim_{n \to \infty } \frac{\frac{n!}{2!\cdot(n-2)!}}{n^{2}+3n-1}=
\lim_{n \to \infty } \frac{\frac{(n-2)!\cdot(n-1)\cdot n}{2\cdot(n-2)!}}{n^{2}+3n-1}= \\ =
\lim_{n \to \infty } \frac{(n-1)\cdot n}{2n^2+6n-2}=
\lim_{n \to \infty } \frac{n^2\left(1-\frac{1}{n}\right)}{n^2\left(2+\frac{6}{n}-\frac{2}{n^2}\right)}=
\frac{1}{2}}\)