trudne lim
Magda: limx−>0+ (x)√x =
limx−>0+ (1/ex−1 − 1/x) =
limx−> −∞ (√x2 +x − x)
4 lut 00:21
Basia:
łatwe !
x
√x = [(
√x)
2]
√x = [
√x√x]
2
t=
√x
lim
x→0+ x
√x = lim
t→0+ (t
t)
2 = 1
2 = 1
| | 1 | | 1 | | 1 | |
limx→0+( |
| −1− |
| ) = |
| − 1 −(+∞) = 1−1−∞=−∞ |
| | ex | | x | | e0 | |
lim
x→−∞(
√x2+x−x} =
lim
x→−∞ (
√x(x+1)−x} =
√−∞*(−∞)−(−
∞) =
√+∞+
∞=+
∞+
∞=+
∞
4 lut 00:34