Ci膮gi, zadanie nr 4874
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
fazi post贸w: 26 |  2015-01-08 11:20:15prosz臋 o rozpisanie a)$\lim_{x \to \infty}\frac{(n+1)(n-3)}{(n+2)(n+3)}$ b)$\lim_{x \to \infty}\frac{(n-1)(n+1)}{n^{2}-n+5}$ c)$\lim_{x \to \infty}(n+1-\frac{1}{n^{2}+3})$ d)$\lim_{x \to \infty}(\frac{n}{n^{2}+1}+n+2)$ |
irena post贸w: 2636 | 2015-01-08 12:34:56 $\frac{n+1)(n-3)}{(n+2)(n+3)}=\frac{n^2-2n-3}{n^2+5n+6}=$ $=\frac{1-\frac{2}{n}-\frac{3}{n^2}}{1+\frac{5}{n}+\frac{6}{n^2}}\to\frac{1}{1}=1$ |
irena post贸w: 2636 | 2015-01-08 12:36:28 b) $\frac{(n-1)(n+1)}{n^2-n+5}=\frac{n^2-1}{n^2-n+5}=\frac{1-\frac{1}{n^2}}{1-\frac{1}{n}+\frac{5}{n^2}}\to1$ |
irena post贸w: 2636 | 2015-01-08 12:37:50 $n+1\to\infty$ $\frac{1}{n^2+3}\to0$ $n+1-\frac{1}{n^2+3}\to[\infty-0]\to\infty$ |
irena post贸w: 2636 | 2015-01-08 12:39:18 $\frac{n}{n^2+2}=\frac{\frac{1}{n}}1+\frac{2}{n^2}}\to0$ $n+2\to\infty$ $\frac{n}{n^2+2}+n+2\to[0+\infty]\to\infty$ |
| strony: 1 | |
Prawo do pisania przys艂uguje tylko zalogowanym u偶ytkownikom. Zaloguj si臋 lub zarejestruj