Помогите решить предел

1) lim 3x²-2/2x²+7

2) lim x²-2x-3/х²-5 х+6

х → 3

1. lim ((3x²-2) / (2x²+7)) = (3*0²-2) / (2*0²+7) = - 2/7

x->0

2. lim ((x²-2x-3) / (x²-5x+6)) = неопределенность [0/0]

x->3

x²-2x-3 = (x+1) * (x-3)

x²-5x+6 = (x-3) * (x-2)

lim (x²-2x-3) / (x²-5x+6) = lim ((x+1) * (x-3) / (x-2) * (x-3) = lim ((x+1) / (x-2)) = (3+1) / (3-2) =

x->3 x->3 x->3

=4/1=4