Let f: R R be a twice differentiable function such that f(2) = 4 and f^ (2) = 0 . If _ x 0 ( 1 + _ e ( f(2+x) 4 ) )^ 1 x^2 = e^3 , then the value of f^ (2) is equal to :

Step-by-step solution

The given limit is of the form 1^ . Let L = _ x 0 ( 1 + _ e ( f(2+x) 4 ) )^ 1 x^2 . Using the standard limit _ x a (1 + g(x))^ 1 h(x) = e^ _ x a g(x) h(x) for 1^ forms: L = e^ _ x 0 _ e ( f(2+x) 4 ) x^2 Let P = _ x 0 _ e (f(2+x)) - _ e (4) x^2 . This is a 0 0 form. Applying L'H 00f4pital's rule: P = _ x 0 f^ (2+x) f(2+x) 2x Since f^ (2) = 0 , this is again a 0 0 form. Applying L'H 00f4pital's rule a second time using the quotient rule: P = _ x 0 f^ (2+x)f(2+x) - (f^ (2+x))^2 (f(2+x))^2 2 Substituting x = 0 : P = f^