Знатоки алгебры Нужно понятное и подробное решение

!Задание на фото.​

Объяснение:

lim(x→3) (x + (x²-9)/(x-3)) =

= lim(x→3) (x) + lim(x→3) (x²-9)/(x-3) =

= [3] + [0/0] =

= 3 + lim(x→3) (x²-9)/(x-3) =

= 3 + lim(x→3) (x-3)*(x+3)/(x-3) =

=3 + lim(x→3) (x+3)=

=3 + [6] =

=3 + 6 = 9

lim(x→3) (x + (x²-9)/(x-3)) =

= lim(x→3) (x) + lim(x→3) (x²-9)/(x-3) =

= [3] + [0/0] =

Воспользуемся правилом Лопиталя:

= 3 + lim(x→3) (x²-9)' / (x-3)' =

= 3 + lim(x→3) (2х+0)/(1+0)=

= 3 + lim(x→3) (2х)=

= 3 + [6]=

=3+6=9

lim(x→-2) ((4-х²)/(х-2) +х) =

=lim(x→-2) (4-х²)/(х-2) + lim(x→-2) (х)=

= [0/-4=0] + [-2]=

=0 + (-2) = -2

lim(x→-2) ((4-х²)/(х-2) +х) =

= lim(x→-2) (4-х²)/(х-2) + lim(x→-2) (х)=

= lim(x→-2) (2-х)*(2+х)/(х-2) + lim(x→-2) (х) =

= lim(x→-2) (-(2+х)) + lim(x→-2) (х) =

= [-(0)=0] + [-2]=

=0 + (-2) = -2