Evaluating an expression when x=infinity

[AStaunton]

can someone please tell me what is the value of the following expression when x=\infty:

-xe^{-x}

my guess is it is either -1 or 0 or -infinity...

 

[HallsofIvy]

First, be careful of your terminology. You cannot "evaluate a function at x= infinity"- infinity is not a number. What you mean is "find the limit as x goes to infinity" (which, itself, is short for "as x gets larger and larger without bound").'

Since, as x "goes to infinity", e^{-x} goes to 0, this is an indeterminant of the form "infinity times 0". It can be put in the form "infinity over infinity" by writing it as -x/e^x and you can then use L'Hopital's rule.

 

[AStaunton]

infinity times 0 = 1 I thought?

I'm thinking this mainly due to the little I know Dirac Delta function..

 

[AStaunton]

after using l'hop's rule:

set f(x)=x g(x)=e^x
and f'(x)=1 g'(x)=e^x

I found f'/g'=1/infinity=0 which implies f/g=0..

Can you please confirm this is correct, as my skills with limits is very limited!

 

[madah12]

you should know that e^x grows very fast and the line y=x grows with slope 1 so it should be easy to picture what lim x--> infinity x/e^x is at x =1000 e^x is about 1.97 *10^434 so this limit is easy to guess

 

[Berko]

Yes, you are correct. But, in general, "infinity" times zero is not 1...it can be anything, which is why we do not allow it.