nick850 said:
no i don't know what that is. from the looks from it on wikipedia it looks pretty complicated. I'm sure it's a better way of doing it but i guess my question was more on how to algebraically manipulate the equation so that the limit isn't indeterminate (0/0)
As I said, if you don't know the Binomial series expansion, then now is the time to learn it. It's just too important to not know it. It's really not that complicated, but it might look bad symbolically, at first.
I don't know if it really is the best method to solve this problem but it does work very quickly in this case. You end up keeping two terms of an infinite series (other terms go to zero with high orders of h). This limiting form basically allows you to do the manipulation you are looking for, and you can easily remove the indeterminate form.
Usually we use L'Hospital's rule in indeterminate cases, but since that involves taking derivatives, you end up with a circular problem where you need to take the derivative of exactly the same form.
Anyway, rather than leave you hanging, I'll reduce the Binomial Series to the end form you would have (with apologies for writing some sloppy math in the interest of motivation). Maybe this will inspire you to bite the bullet and learn it now.
(x+h)^n\approx x^n+nx^{n-1}h when h is small.
If you use this on (x+h)^(-2/5) in the expression ((x+h)^(-2/5)-x^(-2/5))/h you will end up with very simple algebra.
When you get the final answer, you should compare it to the well known derivative rule for powers {{dx^n}\over{dx}}=nx^{n-1}. If fact, you may just want to derive this rule directly and then use it to solve your special case.
