If _ x 0 x^2 + 2x - A (2x) B x (1+x) = 1 2 , where A and B are real constants, then the value of A + B is:

Step-by-step solution

Given limit is _ x 0 x^2 + 2x - A (2x) B x (1+x) = 1 2 Using Maclaurin series expansions: (2x) = 2x - (2x)^3 3! + (1+x) = x - x^2 2 + Substituting these into the limit: _ x 0 x^2 + 2x - A (2x - 8x^3 6 + ) B x (x - x^2 2 + ) _ x 0 (2 - 2A)x + x^2 + 4A 3 x^3 + B x^2 - B 2 x^3 + = 1 2 For the limit to be finite, the coefficient of the lowest power of x in the numerator must be zero (since the denominator starts with x^2 ). 2 - 2A = 0 A = 1 Now, the limit becomes: _ x 0 x^2 + 4 3 x^3 + B x^2 - B 2 x^3 + = 1 B Given tha