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

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.

oh right cancelling! man i feel like an idiot! *shame*

 

lol

 

thanks mezarashi