| 2 | ||
lim(1+ | )2n2 | |
| 3n2 |
| 3*2n+4*4n | ||
lim | ||
| 5*2n−2*5n |
| 2 | 2 | 2 | ||||
(1+ | )2n2=(1+ | )3n2/1.5=((1+ | )3n2)1/1 | |||
| 3n2 | 3n2 | 3n2 |
| 3 (2/5)n+4(4/5)n | 3*0+4*0 | |||
mamy | → | =0 | ||
| 5(2/5)n−2 | 5*0−2 |
| 2 | 1 | |||||||||
limn→∞ (1+ | )2n2 = limn→∞ (1+ | )2n2 = | ||||||||
| 3n2 |
|
| 1 | 1 | |||
= limn→∞ (1+ | )2n2 = limn→∞ (1+ | )1.5n2*21.5 = | ||
| 1.5n2 | 1.5n2 |
| 2 | 1 | |||||||||
Przy przejściu z limx→∞ (1+ | )2n2 do limx→∞ (1+ | )2n2 | ||||||||
| 3n2 |
|
| 2 | ||
podzieliłem licznik i mianownik ułamka | przez 2 | |
| 3n2 |
| 3*2n+4*4n | ||
limn→∞ | = | |
| 5*2n−2*5n |
| 5n(3*(25)n+4*(45)n) | ||
= limn→∞ | = | |
| 5n(5*(25)n−2) |
| 3*(25)n+4*(45)n | 3*0+4*0 | 0 | ||||
= limn→∞ | = | = | = 0 | |||
| 5*(25)n−2 | 5*0−2 | −2 |