obliczyc granice ciagow
daro: a)lim n(√2n2+1 − √2n2−1
15 lis 18:29
00: ∞
15 lis 18:40
Aza:
| | n(√2n2+1−√2n2−1)(√2n2+1+√2n2−1) | |
an = |
| =
|
| | √2n2+1+√2n2−1 | |
| | n(2n2+1 −2n2+1) | |
= |
|
|
| | n(√2+1n2+√2−1n2) | |
n skracamy, i w liczniku redukcja
| | 2 | | 2 | | 1 | | √2 | |
liman = |
| = |
| = |
| = |
|
|
| | √2+√2 | | 2√2 | | √2 | | 2 | |
n→∞
16 lis 00:42
AROB: Spokojnej nocy
Aza, miłych snów.
16 lis 00:55
Aza:
Dobranoc.
16 lis 00:59