Sarahisme Posted June 16, 2005 Share Posted June 16, 2005 i don't see why L'Hop's rule doesnt work by just straight forward application in this limit? the solution i have subs in a value so that the limit has m->0 ? Thanks guys, sorry if this seems like another inane question Link to comment Share on other sites More sharing options...
mezarashi Posted June 16, 2005 Share Posted June 16, 2005 Actually, a direct application of L'Hopital's rule does work, except the substitution of m->0 instead of n->infinity, where m = 1/n is much easier, and you'll be able to get the answer without any of the messy stuff. If you apply L'Hopital's rule, we see here that we have a 0/0. Taking the derivative of the top term, we have [math]\lim_{n\to\infty} \frac{-\frac{1}{n^2}}{ \frac{1}{n^2}e^{-\frac{1}{n}} }[/math] Cancelling out the [math]\frac{1}{n^2}[/math] we then have [math]\lim_{n\to\infty} \frac{-1}{ e^{-\frac{1}{n}} }[/math] which by substituting in n->infinity, will yield -1, the same result as the m substitution. Link to comment Share on other sites More sharing options...
Sarahisme Posted June 16, 2005 Author Share Posted June 16, 2005 oh right cancelling! man i feel like an idiot! *shame* lol thanks mezarashi Link to comment Share on other sites More sharing options...
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