If _ x 0 a 1+2x + b 2x + c e^ -x x^2 = 4 , where a, b, c R , then the cubic equation whose roots are a, b, and c is

Step-by-step solution

Given limit is _ x 0 a 1+2x + b 2x + c e^ -x x^2 = 4 Using Maclaurin series expansions: 1+2x = 1 + 1 2 (2x) + 1 2 (- 1 2 ) 2! (2x)^2 + = 1 + x - x^2 2 + 2x = 1 - (2x)^2 2! + = 1 - 2x^2 + e^ -x = 1 - x + x^2 2! - Numerator = a ( 1 + x - x^2 2 ) + b(1 - 2x^2) + c ( 1 - x + x^2 2 ) = (a + b + c) + (a - c)x + ( - a 2 - 2b + c 2 )x^2 For the limit to exist, the constant term and the coefficient of x must be zero: a + b + c = 0 ... (1) a - c = 0 a = c ... (2) The limit is the coefficient of x^2 : - a 2 - 2b + c 2 = 4 ...