3 − 5 n2 | ||
a) an = | = ; dzielimy licznik i mianownik przez n2 | |
4 − 2n2 |
| ||||||||
= | ||||||||
|
0 − 5 | − 5 | |||
lim an = | = | = 2,5 | ||
0 − 2 | − 2 |
3 | ||
bo lim | = 0 | |
n2 |
4 | ||
i lim | = 0 | |
n2 |
2n +1 | ||
b) an = | ; dzielimy licznik i mianownik przez 2n | |
3 + 4*2n |
| |||||||||||
an = | |||||||||||
|
1 + 0 | 1 | |||
lim an = | = | |||
0 + 4 | 4 |
1 | 3 | |||
lim | = 0 i lim | = 0 | ||
2n | 2n |
(n2 + 3)2 | n4 + 6 n2 + 9 | |||
an = | = | = | ||
(1 −2 n2)2 | 1 − 4 n2 +4 n4 |
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an = | |||||||||||||||||
|
1 + 0 + 0 | 1 | |||
lim an = | = | |||
0 − 0 + 4 | 4 |
n*(n+1) | ||
1 + 2 + 3 + ... + n = | = 0,5(n2 + n) | |
2 |
4 n2 −3 n + 1 | 4n2 −3 n + 1 | |||
an = | = | |||
1+2+3+...+n | 0,5 n2 +0,5 n |
| |||||||||||||||||
an = | |||||||||||||||||
|
4 − 0 + 0 | 4 | |||
lim an = | = | = 8 | ||
0,5 + 0 | 0,5 |
n+2 − ( n +1) | 1 | |||
an = √n+2 − √n+1 = | = | |||
√n+2 + √n+1 | √n+2 + √n+1 |
1 | ||
lim an = [ | ] = 0 | |
∞ +∞ |
a2 − b2 | ||
W takich przykładach korzystamy ze wzoru a − b = | ||
a + b |
1 | ||
(x − 2) + ( x −2)2 + (x − 2)3 + ... = | x2 + 1 | |
4 |
a1 | x − 2 | x − 2 | |||
= | = | ; x ≠ 3 | |||
1 − q | 1 − ( x −2) | 3 − x |
x − 2 | 1 | ||
= | x2 + 1 / * ( 3 − x) | ||
3 − x | 4 |
3 | 1 | |||
x − 2 = | x2 + 3 − | x3 − x / *4 | ||
4 | 4 |