granice
yfahocos: Oblicz granice funkcji
a) limx→0 = x2−12x2−×−1
b) limx→3 = x2−5×+6x2−8×+15
c) limx→0 = √x+1−1×
d) limx→1 = 6|x2−1|
e) limx→−∞ = (3x3 −6x2+7)
f) limx→2+ = x2+5×−3x2−4
7 sty 16:29
Basia: napisz to jakoś porządnie; dotyczy a,b,f
7 sty 20:02
Basia:
ad.c
| √x+1−1 | | (√x+1−1)(√x+1+1) | |
limx→0 |
| = limx→0 |
| = |
| x | | x*(√x+1+1) | |
| x+1−1 | | x | |
limx→0 |
| = imx→0 |
| = |
| x*(√x+1+1) | | x*(√x+1+1) | |
| 1 | | 1 | | 1 | |
imx→0 |
| = |
| = |
| |
| √x+1+1 | | √0+1+1 | | 2 | |
7 sty 20:06
Basia:
ad.d
limx→1 6|x2−1| = 6|1−1| = 6*0 = 0
7 sty 20:07
Basia:
ad.e
| 6 | | 7 | |
limx→−∞ (3x3 −6x2+7) = limx→−∞ x3(3− |
| + |
| ) = |
| x | | x3 | |
−
∞(3−0+0) = −
∞*3 = −
∞
7 sty 20:09