If f (x)= cases 8^x-4^x-2^x+1^x x^2 & , if x>0 e ^x x+i x+ 4 & , if x 0, i R cases is continuous at x=0 , then the value of 500 e^ is

Step-by-step solution

Ensuring continuity at x=0 requires: _ x 0^- f (x) = _ x 0^+ f (x) = f (0) The left-hand limit evaluates to 4 , as e ^x 1 , x 0 , and i x 0 . The right-hand limit transforms to _ x 0^+ (4^x-1)(2^x-1) x^2 . Applying the standard limit _ x 0 a^x-1 x = a yields ( 4)( 2) . Since 4 = 2 2 , this simplifies to 2( 2)^2 . Equating both sides: 4 = 2( 2)^2 . Substituting 4 = 2 2 gives (2 2) = 2( 2)^2 , so = 2 . The expression 500 e ^ becomes 500 e ^ 2 = 500 2 = 1000 . 1000