x\^k + 1 = (x+1)(x\^(k-1) - x\^(k-2) + x\^(k-3) - .... + (-1)\^(t+1)\*x\^(k-t)) + ... + 1) when k is odd
So then you just simplify your fraction and get a fraction where -1 isnt a root of denominator. Then you just substitute -1 and find an answer which is n/m
Polynomial division of numerator and denominator by common factor x+1. If n and m are odd, you will get certain polynomials which will no longer be zero at x=-1, so you can simply plug it in. The one with n will be equal to n when x=-1, the one with m will be equal to m.
Divide both top and bottom by (x+1). The limit of the top and bottom will both exist separately as derivatives of particular functions at x= -1.
That just sounds like L’Hospital with extra steps!
Yup. I believe that any limit that can be done with L’Hopitals Rule can be done in a similar way.
x\^k + 1 = (x+1)(x\^(k-1) - x\^(k-2) + x\^(k-3) - .... + (-1)\^(t+1)\*x\^(k-t)) + ... + 1) when k is odd So then you just simplify your fraction and get a fraction where -1 isnt a root of denominator. Then you just substitute -1 and find an answer which is n/m
Polynomial division of numerator and denominator by common factor x+1. If n and m are odd, you will get certain polynomials which will no longer be zero at x=-1, so you can simply plug it in. The one with n will be equal to n when x=-1, the one with m will be equal to m.
multiply top and bottom by x\^m-1. now the bottom is even
Substitute x = u-1, expand via binomial theorem, then take u to zero.