Объяснение:
1) 1способ:
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
2способ:
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
3) 1способ:
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) + 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]=
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Answers & Comments
Объяснение:
1) 1способ:
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
2способ:
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
3) 1способ:
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
2способ:
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