1. The problem statement, all variables and given/known data Suppose f:(-1,1)→R is three times differentiable on the interval. Assume there is a positive value M so that ⎮f(x)⎮ ≤ M⎮x⎮³ for all x in (-1,1). Prove that f(0)=f'(0)=f"(0)=0. 2. Relevant equations 3. The attempt at a solution My professor started us off, ⎮f(0)⎮≤M(0)=0; f(0)=0 f'(0)=lim as x→0 [(f(x)-f(0))/x-0 = lim as x→0 [f(x)/x]. I know that ⎮f(x)/x⎮≤ 1/⎮x⎮(M⎮x⎮³ ≤ Mx² Which means that f'(0) = 0 I also know that the next step is to find f"(0) = lim as x→0 [(f'(x)-f'(0))/x-0]. I need to know if f'(x) = 1/x?