If f (x)= (27-2 x)^ 1 3 -3 9-3(243+5 x)^ 1 5 , x 0 is continuous at x=0 , then the value of f(0) is

Step-by-step solution

To ensure continuity at x=0 , define f(0) = _ x 0 f(x) where f(x)= (27-2 x)^ 1 3 -3 9-3(243+5 x)^ 1 5 . Direct substitution yields the indeterminate form 0 0 , requiring further analysis. Applying L'Hôpital's rule, differentiate numerator and denominator: N'(x) = - 2 3 (27-2x)^ - 2 3 D'(x) = -3(243+5x)^ - 4 5 The limit becomes _ x 0 N'(x) D'(x) = - 2 3 (27)^ - 2 3 -3(243)^ - 4 5 . Expressing 27=3^3 and 243=3^5 , we simplify: - 2 3 3⁻² -3 3⁻⁴ = - 2 27 - 1 27 = 2 . Alternatively, using the standard limit _ y 0 (a+y)^