Răspuns:
Explicație pas cu pas:
x²+2x-3)²/(x²-1)²=[(x-1)(x-3)]²/[(x-1)(x+1)]²=
[(x-1)(x-3)/(x-1)(x+1)]²=[(x-3)/(x+1)]²
I1=x->1lim[(x-3)/(x+1)]²=[(1-3)/(1+1)]²=(-2/2)²=(-1)²=1
(x²+x-2)²/(x²-3x+2)²=[(x+2)((x-1)/(x-2)/(x -1)]²=
[x+2)/(x-2)]²
L3=x-.>1 lim[(x+2)/(x-2)]²=
[(1+2)/(1-2)]²=(3/-1)²=(-3)²=9
L3=9
__________________________________
b)(x²+2x-3)²/(x-1)²*(x²-3x+2)²*(x²+x-2)²/(x²-3x+2)²=simplifici=
(x²+x-2)²/(x-1)²=[(x+2)(x-1)/(x-1)(x+1)]²=[(x+2)/(x+1)]²
L3=lim[(x+2)/(x+1)]²=[(1+2)/(1+1)]²=(3/2)²=9/4
c)9/4=1*I2*9
9/4=9I2
I2=1/4
