Oblicz granicę funkcji:
a)
| e2x−ex | 1−1 | |||
x→0 | = | = 0 | ||
| 3sin(4x) | 0 |
| (e2x−ex)' | −ex+e2x(2x) | |||
= | = | − i tutaj nie wiem czy | ||
| (3sin(4x))' | 3cos(4)(4x) |
| lnx | 0 | |||
x→0 | = | = 2 | ||
| 2−lnx | 2 |
| (lnx)' | 1 | x | ||||
= | = | * | = 0 ? | |||
| (2−lnx)' | x | 1 |
| ln(1+2x) | 0 | |||
x→0 | = | = 0 | ||
| e−5x−1 | 0 |
| e2x−ex | 0 | |||
limx→0 | = masz symbol [ | ] | ||
| 3sin(4x) | 0 |
| 2e2x−ex | 2−1 | 1 | ||||
=Hlimx→0 | =[ | ]= | ||||
| 3*4cos(4x) | 12 | 12 |
| lnx | 1 | 1 | ||||
2) lim x→0 | =H limx→0{( | ):(− | )}=−1 | |||
| 2−lnx | x | x |
| ln(1+2x) | 0 | |||
c) limx→0 | = symbol [ | ] | ||
| e−5x−1 | 0 |
| −2 | |||||||||
=H limx→0 | = | |||||||||
| −5e−5x | 5 |