Evaluate the limit as x goes to infinity: (x/x+1)^x

Let y = lim (x/x+ 1)^x

then ln(y) = lim x ln(x/x + 1) = lim ln(x/x + 1)/(1/x) this approaches the indeterminate 0/0

Use L'Hopitals rule: ln(y) = lim (1/x(x +1))/(-1/x^2) = lim -x^2/x(x + 1)
as x approaches infinity, the limit approaches -1

so ln(y) = -1 --> y = e^(-1)

The limit as x approaches infinity is 1/e