| 1 | 1 | π | ||||
Mam problem z taką granicą: limx→0− 2x + 1 − | *(arctg( | )+ | ) , czy jest | |||
| x | x | 2 |
? Dzięki!
| arctg(1/x)+π/2 | 0 | |||
limx→0− | masz symbol: [ | ] | ||
| x | 0 |
| 1 | −1 | −1 | ||||
=H limx→0− | * | =limx→0− | =−1 | |||
| 1+(1/x2) | x2 | x2+1 |
| arctg(1/x)+π/2 | ||
limx→0−(2x+1− | )=[0+1+1]=2 | |
| x |
| arctg(1/x)+π/2 | ||
limx→0+(2x+1− | )=−∞ | |
| x |