Answer
$1$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} \cr
& {\text{Evaluating the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = {\infty ^0} \cr
& {\text{This limit has the form }}{\infty ^0}{\text{ }} \cr
& {x^{{e^{ - x}}}} = {e^{^{{e^{ - x}}}\ln x}} = {e^{\frac{{\ln x}}{{{e^x}}}}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{{\ln x}}{{{e^x}}}}} = {e^{{{\mathop {\lim }\limits_{x \to \infty } }^{\frac{{\ln x}}{{{e^x}}}}}}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{e^x}}} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln x} \right]}}{{\frac{d}{{dx}}\left[ {{e^x}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{x{e^x}}} \cr
& L = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^0} = {e^L} = 1 \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = {e^L} \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = 1 \cr} $$