Evaluate the limit using L\'Hospital\'s rule Suppose that Solution As it is 0/0 form Apply L Hospital Rude FDifferentiate Numenator and Denominator w.r.t x , we get lim(x--->0) (11xln11 - 2xln2)/(1) = (110ln11 - 20ln2) = ln(11/2) = 1.7047 ========================================= f\'(x) = 1/(2(4+(5x))*(d/dx)((4+(5x)) = 1/(2(4+(5x))*(1 + 5(1/(2x)).