See.. Since it is continuous at x = 0, so f(0^{-}) = f(0^{+}) = f(0). ------- (X)
Here f(0^{-}) = Lim(x->0) { (1+ax)^{1/x} } = e^{a} .
f(0) = b
and f(0^{+}) =Lim ( x->0) {(x + c)^{1/3} - 1}/ (x)
Now if c = 1 then,
f(0^{+}) =Lim ( x->0) {(x + 1)^{1/3} - 1} /(x) => 0/0
using L hospital rule, we can solve above limit = 1/3 .
Hence from eqn (X) , e^{a} = b = 1/3.
a = ln (1/3)
b = 1/3
c = 1.
Is it okay now ?