Suppose _ x -2 bx^2 + 15x + 15 + b x^2 + x - 2 = L , where b and L are real numbers. Then

Step-by-step solution

For the limit to exist, the numerator must be zero when x = -2 because the denominator is zero at x = -2 . Substituting x = -2 in the numerator: b(-2)^2 + 15(-2) + 15 + b = 0 4b - 30 + 15 + b = 0 5b = 15 b = 3 Substituting b = 3 in the given limit: L = _ x -2 3x^2 + 15x + 18 x^2 + x - 2 L = _ x -2 3(x+2)(x+3) (x+2)(x-1) L = _ x -2 3(x+3) x-1 L = 3(-2+3) -2-1 = 3 -3 = -1 Thus, b = 3 and L = -1 . Answer: b = 3 and L = -1