I have an assignment due Wednesday morning. Right now its 2 AM Tuesday. I was just going to downlload this http://www.maple11.com/applications/app_ce...w.aspx?AID=1808 but I have to register for it.

Anyway, I know how to tackle simple ones like f(x)^g(x) like In(y) = x^2 In(x) but there 1 question that has me baffled when I look at it

x^2 e^x

No one has to answer this as I'm going to look at it again later. I am just asking it andd let it sit before its to late to ask.

It would help if you stated the question rather than just a tiny fragment of it. And, no, I have no intention of becoming a MapleSoft member.

http://img6.imageshack.us/img6/4581/ffffff...ffffffffffx.jpg

That's it and just usee the L Hospital rule.

Registration to maplesoft I think is free, I also have Maple so I'm interested in becoming a member.

So does x^2 e^x become large or small? And what does L'Hospital have to say about it? Well first, L'Hospital says rewrite in terms of f(x)/g(x) where both are small or both are large.

1. Choose f(x) = x^2 , g(x) = e^-x -- both are large
Then look at f'(x)/g'(x) = 2x / -e^-x -- both are large
Then look at f''(x)/g''(x) = 2 / e^-x -- only bottom is large, so x^2 e^x goes to zero.

2. Or Choose f(x) = e^x, g(x) = x^-2 -- both are small
Then look at f'(x)/g'(x) = e^x / -2x^-3 -- both are small
Then look at f''(x)/g''(x) = e^x / 6x^-4 -- both are small
....
You'll never get anywhere here. It was a poor choice.

Ah, you actually answered my question. Sorry I'm a victim of error as much as I errored myself. I've been googling the wrong words It looks like I should have been googling "indeterminate products" if its in the form infinity * 0 I can then express it as a fraction

f(x)g(x)=f(x)/g(x)^-1 or something like that http://math.furman.edu/~mwoodard/math151/docs/sec5_8.pdf

For an indeterminate product. The only answers are 0 or +/- infinity.

http://video.google.com/videoplay?docid=1820106163263850864

Total shot in the dark here but is the reason Mathematica:

http://img4.imageshack.us/img4/6262/mathhhhh.jpg


likes to give me the limit as 0


Because e^x, where x is -Infinity approaches 0 faster than any other power x^n http://www.math.tamu.edu/~fulling/coalweb/lhop.htm

And since I only have -Infinity^1 in the numerator, its too slow to not swallowed by e^-Infinity in the denominator?

Exponentials beat powers. That's the rule.

If you have x^n exp(-x) then view it as x^n / exp(x), expand exp(x) in a power series and you'll see you get x^m terms where m>n and they'll dominate x^n at large x, thus sending x^n exp(-x) to zero as x->infty.

Generalisations are immediate and trivial.