Let r be defined on V (x0) = {x : |x − x0| < a}. If r is n-times differentiable at x0 and r(x0) = r'(x0) = r''(x0) = · · · = r(n)(x0) = 0 (nth derivative),

prove that the limit as x approaches x0 of r(x)/[(x-x0)^n] = 0.

I use Taylor's Theorem, but all of the terms go away since the r(x0)'s all equal zero. It seems I'm left only with the remainder:

r(n+1)(c) * (x-x0) / (n+1)! . Somehow I need to show that the limit of the remainder over (x-x0)^n+1 is zero.

I'm trying to prove it by induction. I can show the base case easily. If I assume the claim is true for all n, then I'm left showing that the limit of the Lagrange Remainder as x goes to x0

r(n+1) (c) *(x-x0)^(n+1)

----------------------------

(n+1)!

-------------------------

(x-x0)^(n+1)

equals zero right? The r(n+1) (c) is the n+1th derivative of r at a point c between x0 and x. How do I do this? I'm pretty sure this problem needs to incorporate the remainder somehow.

If I assume the claim for (n-1) and try to show that the claim holds for n, it seems trivial since r(n) (x0) = 0. I'm seriously missing something.