# Thread: Fairly simple limit proof?

1. ## Fairly simple limit proof?

How would I prove that

$\displaystyle \lim\limits_{x\to\infty}xe^{-x}=0$ ?

I've only done eps-delta proofs for finite limit points, not to infinity.

2. Are you familiar with L'Hopitals rule?

3. Yes. I think I may know how to solve it then. -- Wait, but LHopitals rule is for finding the limit, not proving it?

4. Originally Posted by scorpion007
Yes. I think I may know how to solve it then. -- Wait, but LHopitals rule is for finding the limit, not proving it?
Surely if you find that the limit is equal to zero by using a valid theorem like l'Hopital's Rule, then you have proved that the limit is equal to zero! But context might be important ..... Where has the question come from?

5. It arose indirectly from me solving an integral of the form $\displaystyle \int_0^\infty xe^{-x}$ where I had to evaluate the limit of a similar expression, and was curious if I could prove it using a method like e-d proofs.

6. Originally Posted by scorpion007
It arose indirectly from me solving an integral of the form $\displaystyle \int_0^\infty xe^{-x}$ where I had to evaluate the limit of a similar expression, and was curious if I could prove it using a method like e-d proofs.
In that case, why use a cannon when a flyswatter will do?

7. Only because I was curious if I could prove it -- not because it was strictly necessary to do so.

8. Originally Posted by scorpion007
How would I prove that

$\displaystyle \lim\limits_{x\to\infty}xe^{-x}=0$ ?

I've only done eps-delta proofs for finite limit points, not to infinity.

A some kind of try:

By definition...

For every $\displaystyle \epsilon>0$ exist $\displaystyle x>1$ so that:

$\displaystyle |xe^{-x}|<\frac{1}{x}<\epsilon$