Derivative at a is 0 for any order n

[celtics2004]

Homework Statement

Let U be an open interval in R, let f : U $$\rightarrow$$ R be infinitely differentiable and let a $$\in$$ U. Prove that if f has a zero arbitrarily close to a then f$$^{(n)}$$(a) = 0 for all n $$\geq$$ 0.

Homework Equations

The Attempt at a Solution

f is differentiable so f is continuous. f is continuous so for any $$\epsilon$$ > 0 there is a $$\delta$$ > 0 such that when | x - a | < $$\delta$$ then | f(x) - f(a) | < $$\epsilon$$. there is an x where | x - a | < $$\delta$$ such that f(x) = 0. Therefore | f(a) | < $$\epsilon$$. But epsilon is abitrary so f(a) = 0.

I'm not sure how to approach proving that the nth derivative of f is 0 at a. Any thoughts?

 

[Dick]

You could start by showing that f'(a)=0 using pretty much the same argument with the difference quotient. To get to higher derivatives, start using the mean value theorem.