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.
The Attempt at a Solution
My professor started us off,
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
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?