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Originally posted by sonhouse
What does that do, if anything, to the original answer of $7500?

Nice bit of analysis for sure.

It doesn't do anything to the original problem. Your post with X/0 made me curious what the limit was as X approached 1. I discovered that it was n-2, and then spent an insane amount of time figuring out why.

Originally posted by SwissGambit
It doesn't do anything to the original problem. Your post with X/0 made me curious what the limit was as X approached 1. I discovered that it was n-2, and then spent an insane amount of time figuring out why.

L'Hopital's rule to the rescue!

lim_{x->1}f(x)/g(x) = lim_{x->1}f'(x)/g'(x) = lim_{x->1}(1-(n-1)x^{n-2})/-1 = n-2

Originally posted by Agerg
L'Hopital's rule to the rescue!

lim_{x->1}f(x)/g(x) = lim_{x->1}f'(x)/g'(x) = lim_{x->1}(1-(n-1)x^{n-2})/-1 = n-2

L'Hopi-who?

And my calculus is far too rusty to follow what you just did.

Originally posted by SwissGambit
L'Hopi-who?

And my calculus is far too rusty to follow what you just did.

http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

For what its worth, it's well known that for all integers n, 1-x^n = (1+x+x^2+...+x^{n-1})(1-x); so your approach could have been written simply as follows:
(x-x^{n-1})(1-x) = x(1-x^{n-2})/(1-x)
= x(1+x+x^2+...+x^{n-3})(1-x)/(1-x)

by applying the limit as x tends to 1 this becomes 1(1+1+ ...(n-2 times in total) + 1) = n-2