| 5x2+3x−1 | ||
f(x)= | ||
| x−1 |
| f(x) | ||
gdzie a = lim | ||
| x |
| 5*x2+3*x−1 | ||
a = lim | dzielimy licznik i mianownik przez x2 | |
| x*(x−1) |
| 5 + 3/x − 1/x2 | ||
a = lim | = 5 | |
| 1 − 1/x |
| 5*x2+3*x−1 | 5*x2+3*x−1−5*x*(x−1) | |||
b = lim( | − 5*x) = lim | = | ||
| x−1 | x−1 |
| 8*x−1 | 8 − 1/x | |||
lim | = lim | = 8 | ||
| x−1 | 1 − 1/x |
| f(x) | f(x) | |||
tak , bo lim | = lim | |||
| x | x |