Question

1)Lim (x^3+x-5) x→3 2)lim (2x^3-5x^2+x-4) x→-1 3lim {(7x+2(4x-3)(5x+1)} x→1

Answer 1

Ответ:

1) 25

2) -12

3) 19

Пошаговое объяснение:

1) Lim (x³+x-5)=3³+3-5=37-2=25

  x→3

2) lim (2·x³-5·x²+x-4)=2·(-1)³-5·(-1)²+(-1)-4= -2-5-1-4= -12

    x→-1

3) lim (7·x+2·(4·x-3)·(5·x+1))  = 7·1+2·(4·1-3)·(5·1+1)=

   x→1

=7+2·(4-3)·(5+1)=7+2·1·6=7+12=19

Answer 2

To find the limits of the given expressions, we can substitute the values of 'x' into the expressions and evaluate them.

1. lim (x^3 + x - 5) as x approaches 3: Let's substitute x = 3 into the expression: lim (x^3 + x - 5) as x → 3 = 3^3 + 3 - 5 = 27 + 3 - 5 = 25

Therefore, lim (x^3 + x - 5) as x approaches 3 is equal to 25.

2. lim (2x^3 - 5x^2 + x - 4) as x approaches -1: Let's substitute x = -1 into the expression: lim (2x^3 - 5x^2 + x - 4) as x → -1 = 2(-1)^3 - 5(-1)^2 + (-1) - 4 = 2(-1) - 5(1) - 1 - 4 = -2 - 5 - 1 - 4 = -12

Therefore, lim (2x^3 - 5x^2 + x - 4) as x approaches -1 is equal to -12.

3. lim {(7x + 2(4x - 3)(5x + 1)} as x approaches 1: Let's substitute x = 1 into the expression: lim {(7x + 2(4x - 3)(5x + 1)} as x → 1 = {7(1) + 2(4(1) - 3)(5(1) + 1)} = {7 + 2(4 - 3)(5 + 1)} = {7 + 2(1)(6)} = {7 + 2(6)} = {7 + 12} = 19

Therefore, lim {(7x + 2(4x - 3)(5x + 1)} as x approaches 1 is equal to 19.