# Thread: Infinitely differentiable function question

1. ## Infinitely differentiable function question

Suppose f:R -> R is infinitely differentiable with f(0) = 0. Prove that all derivatives of f at 0 are 0 if and only if $\lim_{x \to 0} = \dfrac{f(x)}{x^n} = 0$

I get the general idea and can prove tex] \lim_{x \to 0} = \dfrac{f(x)}{x^n} = 0 \implies f^{(n)} = 0 [/tex] but can't seem to get the other direction of the proof.

I suppose I have to find the limit definition of $f^{(n)}(0)$ and then simplify it to get the required limit but I can't seem to get the definition.

EDIT: Would it be ok to say
$f''(0) = \lim_{x \to 0} \dfrac{f'(x)}{x} = \lim_{x \to 0} \dfrac{\lim_{x \to 0} \dfrac{f(x)}{x}}{x} = \lim_{x \to 0} \dfrac{f(x)}{x^2}$
for example and then prove the formula for the nth derivative by induction

2. Originally Posted by cana
Would it be ok to say
$f''(0) = \lim_{x \to 0} \dfrac{f'(x)}{x} = \lim_{x \to 0} \dfrac{\lim_{x \to 0} \dfrac{f(x)}{x}}{x} = \lim_{x \to 0} \dfrac{f(x)}{x^2}$
for example and then prove the formula for the nth derivative by induction
No, because $f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}h$ (you can't use the same letter).
You can apply the Taylor's theorem.

3. Surely using the other definition $f'(c) = \lim_{x \to c} \frac{f(x)-f(c)}{x-c}$ would make that fine?

### proof C[a,b] and C[0,1] are isometric

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