Formal proof for $\lim_{x\to\infty}x\exp(-x) =0$

Question

I intuitively understand that

$$\lim_{x\rightarrow\infty} xe^{-x}=0$$

as the $e^{-\infty}$ approaches zero faster than $x$ approaches infinity. But this requires one to have a knowledge of the property of exponents. Is there any way to prove this formally that is mathematically sound? For example, how would the person without the knowledge of the property of exponents can derive the same answer?

Answer 1

Recall that $e^x = 1 + x + \dfrac{x^2}{2!} + \cdots$. Hence, $e^x > \dfrac{x^2}2 \implies xe^{-x} < \dfrac2x$. Now conclude what you want.

Answer 2

A quick suggestion (that should be a comment): try using the series expansion of $e^{-x}$.