24 lis 20:43
Aneta20: up
24 lis 22:16
janek191:
1.
| 3 | | 3 | |
f(x) = ( 1 + |
| )4*[ (1 + |
| )2x]1/2 |
| 2x | | 2x | |
więc
lim f(x) = 1*[ e
3]
0,5 = e
1,5
x→
∞
24 lis 23:54
janek191:
5. lim 31x−2 = +∞
x→ 2+
25 lis 00:00
janek191:
6.
| √1 + x2 − 1 | | 1 + x2 − 1 | | x | |
f(x) = |
| = |
| = |
| |
| x | | x*( √1 + x2 + 1) | | √1 + x2 + 1 | |
więc
x→0
25 lis 00:04
janek191:
Masz odpowiedzi ?
25 lis 00:27
Aneta20: Dziękuję, niestety nie.
25 lis 00:47