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