lim x-> 0 (1+ax)^(1/x)
lim x-> 0 e^ { ln (1+ax)^ (1/x)
lim x-> 0 e^ {1/x ln (1 +ax) }
lim x-> 0 e^ { ln(1 +ax)/x }
Using limit properties,
e^ { lim x-> 0 ln(1+ax)/x }
Applying L'hospitals rule we get:
e^ [ lim x-> 0 d{ln(1+ax)}/dx / d {x}/dx ]
e^ { lim x-> 0 a/(1+ax) }
e^ { a/(1+0)}
e^a