гдз 10 клас математика Істер 2018 вправа 17.16
10 клас ➠ математика ➠ Істер
Вправа 17.16
Умова:
Обчисліть границі:
1) lim (x2-x-6)/(x+2);
х→-2
2) lim (x2-1)/(x2+5x+4);
х→-1
3) lim (x3-8)/(x-2);
х→2
4) lim (1-x)/(1-√х).
х→1
Відповідь:
1) lim (x2-x-6)/(x+2) = (0/0) = lim ((x+2)(x-3))/(x+2) =
х→-2 х→-2
= lim (x - 3) = -2 - 3 = -5
х→-2
x2 - x - 6 = 0
x1 + x2 = 1 x1 = -2
{ {
x1 - x2 = -6 x2 = 3;
2) lim (x2-1)/(x2+5x+4) = (0/0) = lim ((x-1)(x+1))/((x+1)(x-1)) =
х→-1 х→-1
= lim (x-1)/(x+4) = -2/3 = -2/3
х→-1
x2 + 5x + 4 = 0
x1 + x2 = -5 x1 = -1
{ {
x1 - x2 = 4 x2 = -4;
3) lim (x3-8)/(x-2) = (0/0) = lim ((x-2)(x2+2x+4))/(x-2) =
х→2 х→2
= lim (x2 + 2x + 4) = 4 + 4 + 4 = 12;
х→2
4) lim (1-x)/(1-√х) = (0/0) = lim ((1-√х)(1+√х))/(1-√х) =
х→1 х→1
= lim (1 + √х) = 1 + 1 = 2.
х→1
Умова:
Обчисліть границі:
1) lim (x2-x-6)/(x+2);
х→-2
2) lim (x2-1)/(x2+5x+4);
х→-1
3) lim (x3-8)/(x-2);
х→2
4) lim (1-x)/(1-√х).
х→1
Відповідь:
1) lim (x2-x-6)/(x+2) = (0/0) = lim ((x+2)(x-3))/(x+2) =
х→-2 х→-2
= lim (x - 3) = -2 - 3 = -5
х→-2
x2 - x - 6 = 0
x1 + x2 = 1 x1 = -2
{ {
x1 - x2 = -6 x2 = 3;
2) lim (x2-1)/(x2+5x+4) = (0/0) = lim ((x-1)(x+1))/((x+1)(x-1)) =
х→-1 х→-1
= lim (x-1)/(x+4) = -2/3 = -2/3
х→-1
x2 + 5x + 4 = 0
x1 + x2 = -5 x1 = -1
{ {
x1 - x2 = 4 x2 = -4;
3) lim (x3-8)/(x-2) = (0/0) = lim ((x-2)(x2+2x+4))/(x-2) =
х→2 х→2
= lim (x2 + 2x + 4) = 4 + 4 + 4 = 12;
х→2
4) lim (1-x)/(1-√х) = (0/0) = lim ((1-√х)(1+√х))/(1-√х) =
х→1 х→1
= lim (1 + √х) = 1 + 1 = 2.
х→1
