Horizontal asymptote via L'Hopital's Rule

Rewrite $s(x) = x e^{-x} = \dfrac{x}{e^{x}}$. As $x \to \infty$, the numerator $x \to \infty$ and the denominator $e^{x} \to \infty$, so this is the indeterminate form $\infty/\infty$, which justifies applying L'Hopital's Rule.

Differentiating numerator and denominator separately gives $\lim_{x\to\infty} \dfrac{x}{e^{x}} = \lim_{x\to\infty} \dfrac{1}{e^{x}} = 0$. Since this limit is finite, the line $y = 0$ is a horizontal asymptote of $s(x)$ as $x \to \infty$.

Using the product rule, $s'(x) = e^{-x} - xe^{-x} = e^{-x}(1-x)$. Setting $s'(x)=0$ gives $e^{-x}(1-x)=0$; since $e^{-x}$ is never zero, the only critical point is $x=1$.

For $x<1$, $1-x>0$ so $s'(x)>0$ (increasing), and for $x>1$, $1-x<0$ so $s'(x)<0$ (decreasing). Thus $s(x)$ changes from increasing to decreasing at $x=1$, giving a local maximum there, with value $s(1) = 1\cdot e^{-1} = e^{-1}$.