гдз 10 клас математика Істер 2018 вправа 17.15
10 клас ➠ математика ➠ Істер
Вправа 17.15
Умова:
Обчисліть границі:
1) lim (x-2)/(x2+2x-8);
х→2
2) lim (x2+x-3)/(x2-1);
х→1
3) lim (x3+1)/(x+1);
х→1
4) lim (3-√x)/(9-х).
х→
Відповідь:
1) lim (x-2)/(x2+2x-8) = (0/0) = lim (x-2)/((x-2)(x+4)) =
х→2 х→2
= lim 1/(x+4) = 1/6
х→2
x2 + 2x - 8 = 0
x1 + x2 = -2 x1 = 2
{ {
x1 - x2 = -8 x2 = -4;
2) lim (x2+x-3)/(x2-1) = (0/0) = lim ((x-1)(x+3))/((x-1)(x+1)) =
х→1 х→1
= lim (x+3)/(x+1) = 4/2 = 2
х→1
x2 + 2x - 3 = 0
x1 + x2 = -2 x1 = 1
{ {
x1 - x2 = -3 x2 = -3;
3) lim (x3+1)/(x+1) = (0/0) = lim ((x+1)(x2-x+1))/(x+1) =
х→1 х→1
= lim (x2 - x + 1) = 1 + 1 + 1 = 3;
х→1
4) lim (3-√x)/(9-х) = (0/0) = lim ((3-√х)/(3-√х)(3+√х) =
х→9 х→9
= lim 1/(3+√х) = 1/6.
х→9
Умова:
Обчисліть границі:
1) lim (x-2)/(x2+2x-8);
х→2
2) lim (x2+x-3)/(x2-1);
х→1
3) lim (x3+1)/(x+1);
х→1
4) lim (3-√x)/(9-х).
х→
Відповідь:
1) lim (x-2)/(x2+2x-8) = (0/0) = lim (x-2)/((x-2)(x+4)) =
х→2 х→2
= lim 1/(x+4) = 1/6
х→2
x2 + 2x - 8 = 0
x1 + x2 = -2 x1 = 2
{ {
x1 - x2 = -8 x2 = -4;
2) lim (x2+x-3)/(x2-1) = (0/0) = lim ((x-1)(x+3))/((x-1)(x+1)) =
х→1 х→1
= lim (x+3)/(x+1) = 4/2 = 2
х→1
x2 + 2x - 3 = 0
x1 + x2 = -2 x1 = 1
{ {
x1 - x2 = -3 x2 = -3;
3) lim (x3+1)/(x+1) = (0/0) = lim ((x+1)(x2-x+1))/(x+1) =
х→1 х→1
= lim (x2 - x + 1) = 1 + 1 + 1 = 3;
х→1
4) lim (3-√x)/(9-х) = (0/0) = lim ((3-√х)/(3-√х)(3+√х) =
х→9 х→9
= lim 1/(3+√х) = 1/6.
х→9