Help me, stupid math ruining my life!

Ok here's the deal, I have to submit homework for tommorow and I can't solve this:

What is the limit of:
x--> - infinity     X * e^(-X)
Nothing works for me :(   how do I do it?

Oh, sorry I didn't use the commands.
[math]X e^(-X)[\math]

I'm new to this, sorry if it didn't come up right.

x/e^x < x/(1+x + x^2/2) -> 0 as x -> inf

Ok thanks. How did you come up with that way? How do I find the inspiration to get result/ways whatever?


Also troubles me:
x^5x     x-->0+
and
(cos x)^(1/x)    x-->0

Sorry for typos and thanks a lot.

Well you know e^x is 1 + x + x^2/2!... by definition, so it's not too hard to show that it grows faster than and "overpowers" any polynomial.

For limits of functions, sequences and series, comparation is a common tool.

x^5x =  e^(5x ln x)  then the problem reduces to finding the limit of 5x ln x as x -> 0. (Try changing variables to x = e^y)

(cos x)^(1/x) = (1- x^2/2 + O(x^4))^(1/x)
 = 1 - 1/x (x^2/2 + O(x^4)) - 1/x (1/x -1)(x^2 + O(x^4)^2 ...
=  1 - x/2 + O(x^2)
which tends to 1  (look up big O notation if you're unfamiliar with it)

oops, minor typos


(cos x)^(1/x) = (1- x^2/2 + O(x^4))^(1/x)
 = 1 - 1/x (x^2/2 + O(x^4)) + 1/x (1/x -1)(x^2 + O(x^4))^2 ...
=  1 - x/2 + O(x^2)

Thanks a lot!

Oh wait a sec,
I still don't know how I calculate 
5*X ln X
How do I do that?

Thanks ahead.

Let x = e^y

5x ln x = 5y e^y

As x -> 0, y -> -inf

Letting y-> -inf,  5y e^y -> 0
So 5x ln x -> 0 as x -> 0

So x^5x = e^(5x ln x) -> e^0 = 1

Awesome! Thanks a lot!