Escaping L'Hopital's Loop

Date: 10/14/2003 at 02:15:07
From: Derek 
Subject: L'Hopital's rule

Given 

  F(X) = e^(-1/x^2) when x does not = 0

       = 0          when x = 0

use L'Hopital's rule to show that F'(0) = 0.

The problem is that when I find the derivative of the function I get

  F'(x) = -[e^(-1/x^2)] / x^2 

which gives 0/0 when you put in x=0.  So I use L'Hopital's rule here.
 But when I do that I end up with 

  F'(x)= [-2 e^(-1/x^2)] / x^3
 
which is still 0/0.  I am not sure how to proceed, because each time I
use the rule I keep getting 0/0.  The exponent in the denominator gets
bigger, but that doesn't change anything. 

Any help would be greatly appreciated!

Date: 10/14/2003 at 06:32:04
From: Doctor Luis
Subject: Re: L'Hopital's rule

Hi Derek,

The good thing about L'Hopital is that it doesn't have to be used on
the indeterminate form 0/0. It can also be used on +oo/+oo.

We can make use of that by rewriting the fraction a/b as (1/b)/(1/a).
Inspired by this, we write

  2e^(-1/x^2) = (2/x^3) / e^(1/x^2)

where I've used the fact that 1/e^(-1/x^2) = e^(+1/x^2).  Now you can
happily apply L'Hopital's rule to find your limit.  Actually, I had to
apply it twice, using the same trick, but it works.

Does this help?  Let me know how it turns out.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/14/2003 at 16:51:27
From: Derek 
Subject: Thank you (L'Hopital's rule)

Thank you!  Your speedy reply was extremely helpful, and very easy to
understand.