| √4x | ||
f= | ||
| √x−5 |
pionowa: x=5
pozioma x=2 − ale jak to wynaczyć? 
| 2√x | 2√x | |||
lim f(x) = lim | = lim | = 2 | ||
| √ x (1 − 5/x ) | √x |
| 5 | |
→0 przy x→∞ | |
| x |
| y | ||
gdzie a = lim(x→∞) | , b = lim(x→∞)(y − a*x) | |
| x |
| 4*x | ||
y = √ | ||
| x − 5 |
| y | 1 | 4*x | ||||
m = lim(x→∞) | = lim(x→∞) | *√ | ||||
| x | x | x − 5 |
| 4 | 2 | |||
m = lim(x→∞)√ | = | = 0 | ||
| x*(x − 5) | ∞ |
| 4*x | 4*x | |||
n = lim(x→∞)( √ | − 0*x) = lim(x→∞) √ | |||
| x − 5 | x − 5 |
| 4 | 4 | |||
n = lim(x→∞)√ | = √ | = 2 | ||
| 1 − 5/x | 1−0 |