# Limit and partial derivatives proof

1. Dec 19, 2007

### cezarion

1. The problem statement, all variables and given/known data

Prove that if all partial derivatives up to order $$n$$ are zero at $$\vec{x}$$ and $$f(x) = 0$$ then $$\displaystyle\lim_{h \rightarrow 0} \dfrac{f(x + h)}{|h|^n} = 0$$

2. Relevant equations

$$\displaystyle\lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{|h|} = 0$$
$$f(x) = 0$$

3. The attempt at a solution

I am not sure if I should be doing this by induction. The first case is trivial. For the general case I am unable to move beyond either the limit $$\displaystyle\lim_{h \rightarrow 0} \dfrac{0}{|h|^{k}}$$. I am also not sure how to formulate the equation for the k-th partial derivative. Is it permissible to use the same $$h$$ for both a given derivative and the next derivative? Thanks for your help.

2. Dec 19, 2007

### HallsofIvy

Staff Emeritus
Are you certain that the first case it "trivial"? Knowing that the partial derivatives are tells you that $$\lim_{h\rightarrow 0}\frac{f(\vec{x}+ h\vec{e_i})- f(\vec{x})}{h} = \lim_{h\rightarrow 0}\frac{f(\vec{x}+h\vec{e_i})}{h}= 0$$ where h is a real number and $\vec{e_i}$ is the unit vector in the ith coordinate direction. What you want to prove, $$\lim_{\vec{h}\rightarrow \vec{0}}\frac{f(\vec{x}+ \vec{h})}{|\vec{h}|^n}= 0$$ requires that h be a general vector going to the 0 vector. (It is, of course, very easy to prove one from the other. What is the derivative in direction $\vec{h}$ in terms of the gradient or partial derivatives?)

3. Dec 19, 2007

### cezarion

Thank you for the reply. This was actually proved in our textbook, so I am just citing the result in the proof for this problem, leaving just the general case that needs to be proved.