Answer
$0$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} x e^{-x}=\lim\limits_{x \to +\infty} \dfrac{x}{e^x}$.
But $\lim\limits_{x \to +\infty} \dfrac{x}{e^x}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to +\infty} \dfrac{ \dfrac{d}{dx} x}{\dfrac{d}{dx} e^x}=\lim\limits_{x \to +\infty} \dfrac{1}{e^x} \\ =\dfrac{1}{\infty}\\=0$
