Look right.

Then

If a,b>0lim x -> 0 (1+a/x)^bx, this one is slightly harder i can get the first derivative then i'm not sure what to do. So for the first derivative I have b(1+a/x)^bx-1*(a/x^2) then i'm lost. {edit} ok i just realized different form g(x) ln f(x) so if i do that i get b * ( -a/x^2) so -ab/x^2 would that be negative infinite then?

(1+ax)^(bx)=exp(ln(1+ax)^(bx)) for x>0.

But then,

exp(bx*ln(1+ax))

Note that,

bx*ln(1+ax) = ln(1+ax)/(1/bx)

You can use L-Hopital Rule on this.