| 1 | 1 | |||
lim dla x→0 ( | − | ) | ||
| x | arctgx |
| 1 | 1 | x | arctgx−x | ||||
− | = | * | |||||
| x | arctgx | arctgx | x2 |
| arctgx−x | 1/(1+x2)−1 | −x | ||||
limx→0 | = H = limx→0 | = limx→0 | = | |||
| x2 | 2x | 2+2x2 |
| 1 | 1 | ||
− | →0 | ||
| x | arctgx |
| x | ||
A co z granicą ułamka | ? To jest 0 przez 0. Czy można to pominąć? | |
| arctgx |
| x | |
→1 przy x→0 | |
| arctgx |
| arctgx−x | ||
lim przy x dążącym do 0 | ||
| x2 |
| x − arctg x | ||
To policz teraz limx→0 | . | |
| x3 |