# Fairly simple limit proof?

• Jun 27th 2010, 03:57 AM
scorpion007
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.
• Jun 27th 2010, 04:18 AM
Are you familiar with L'Hopitals rule?
• Jun 27th 2010, 04:29 AM
scorpion007
Yes. I think I may know how to solve it then. -- Wait, but LHopitals rule is for finding the limit, not proving it?
• Jun 27th 2010, 05:20 AM
mr fantastic
Quote:

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?
• Jun 27th 2010, 04:18 PM
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.
• Jun 27th 2010, 05:09 PM
Plato
Quote:

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?
• Jun 27th 2010, 05:19 PM
scorpion007
Only because I was curious if I could prove it -- not because it was strictly necessary to do so.
• Jun 27th 2010, 05:28 PM
Also sprach Zarathustra
Quote:

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:

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