use identity
[tex]\lim f(x) = \lim e^{ln f(x)} = e^{\lim ln f(x)}[/tex]
The limit now is:
[tex]e^{\lim ln x^{e^{-x}}} \\ \\ =e^{\lim e^{-x} ln x} \\ \\ =e^{\lim \frac{ln x}{e^x}} \\ \\ =e^0 = 1[/tex]
The limit of lnx/e^x is 0 because e^x is exponentially greater than ln(x) as x
goes to infinity. You could also use L'Hopitals rule here to obtain the limit of 0.
