Escaping L'Hopital's LoopDate: 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. |
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