Prove that $\lim_{x \rightarrow \infty} x^ne^{-x}=0$ I'm trying to prove that $$\lim_{x \rightarrow \infty} x^ne^{-x}=0$$

So far I have said:

Let $\epsilon&gt;0.$ 

Then $$|x^ne^{-x}-0|=|x^ne^{-x}| \leq |x^n...|$$

I'm not sure how to bound $e^{-x}$ though. How can I do this?
Don't try to prove it by definition if you don't have to.

Since $n$ is a fixed number in this limit we can apply L' Hospital rule n times.

$$
\lim_{x \rightarrow +\infty} x^ne^{-x}=\lim_{x \rightarrow +\infty} \frac{x^n}{e^{x}}=\lim_{x\rightarrow +\infty}\frac{n!}{e^x}=0
$$

Even though here it wasn't necessary note that since $x\rightarrow +\infty$ we can observe $x\ge n_0 , n_0\in \mathbb{N}$
Do expansion of e^x. You may know it. Then use either Mathematical Induction or Reasoning to prove that limit equals 0 ... but I prefer to apply L'Hospital's Rule, because that will prevent you from doing Hard Reasoning.