Let f: R R be a continuous function satisfying _ x 0 f(x) - 1 x = 4 . If L = _ x 0 ( 1 x ₀^ x f(t) dt )^ 3 x , then the value of (L) is :

Step-by-step solution

From the given condition _ x 0 f(x) - 1 x = 4 , the limit exists and is finite. Since the denominator approaches 0 , the numerator must also approach 0 . Thus, f(0) = 1 . Now, evaluate the base of the limit L as x 0 : _ x 0 1 x ₀^ x f(t) dt This is a 0 0 form. Using L'Hopital's rule and the Leibniz rule: _ x 0 f(x) 1 = f(0) = 1 Since the base approaches 1 and the exponent approaches , L is of the 1^ form. Using the formula for 1^ limits: L = ( _ x 0 3 x ( ₀^ x f(t) dt x - 1 ) ) L = ( _ x 0 3 ( ₀^ x f(t) dt - x x^2