Lim as x approaches infinity (1+3/x)^(5x)

1 because as x approach infinity the 3/x goes to zero which just leaves 1^(infinity) and 1 to any power is 1.
they show that lim x-> infinity (1+a/x)^x=e^a
so if we have lim x-> infinity (1+3/x)^x=e^3
and if we have lim x-> infinity (1+3/x)^(5x) = lim x->infinity [(1+3/x)^x]^5
(the inside goes to e^3) = e^[3(5)]=e^15