Little-o, Big-O and differentiation The functions $f,g, h: \mathbb{R} \rightarrow \mathbb{R}$ have the origin 0 as an internal point of their domain.


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*Prove that if $f = \mathcal{O}(x^{k})$, $f = \mathcal{o}(x^{k-1})$

*Prove that if $g = \mathcal{o}(x)$, g is differentiable in 0. Next assume that for $m \in \mathbb{N}_{\leq 1}$, $g = \mathcal{o}(x^{m})$ and $g$ is $m$ times differentiable. Calculate $g^{(k)}(0)$ for $k \in \{0, 1, ..., m\}$

*Assume $h$ is $n+1$ times differentiable with $h^{(n+1)}$ continuous and $h = \mathcal{o}(x^{n})$. Prove that $h = \mathcal{O}(x^{n+1})$.



What I have tried is the following.



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*As $f = \mathcal{O}(x^{k})$, there exists an open interval $I \subset Dom(f)$ with $0 \in I$ and $C>0$ so that if $x \in I$, $0 \leq |f(x)| \leq C |x|^{k}$. Using the squeeze theorem it follows that $\lim_{x \rightarrow 0} f(x) = 0$, and because $0$ is an internal point of the domain of $f$, $f(0)=0$ so that the first condition for $f$ to be an element of $\mathcal{o}(x^{k-1})$ has been met. Next observe that $I = ]a,b[$ for some $a, b \in \mathbb{R}$ with $a < b$. Because $0 \in I$, $a < 0 < b$. Let $\delta < min\{|a|,|b|,1\}$ and name $D = ]-\delta,\delta[$. It follows that $0 \in D \subset I$. For $x \in D$, $|x| < 1$ and $|x|^{k} < |x|^{k-1}$, so $|h(x)| \leq C|x|^{k} \leq C|x|^{k-1}$.


Is this the right direction to go in? Because I cant find the answer from here.


*Because $g = \mathcal{o}(x)$, g(0) = 0 and $lim_{x \rightarrow 0} \frac{g(x)}{|x|} = 0$. Then $lim_{x \rightarrow 0} \frac{g(x)}{|x|} = lim_{x \rightarrow 0} \frac{g(x)-g(x)}{|x|-0} = lim_{x \rightarrow 0} \frac{g(x)-g(x)}{x-0} = g'(0) = 0$.


Doesnt it follow from this that every higher derivative of $g$ also equals 0?


*Nothing much, cause I have no idea how to start.



Any help is greatly appreciated.

edit:
The definitions used are:


*

*$f = \mathcal{o}(x^{k}) \leftrightarrow f(0) = 0, \lim_{x \rightarrow 0} \frac{f(x)}{|x|^{k}} = 0$

*$f = \mathcal{O}(x^{k}) \leftrightarrow \forall x \in I, |f(x)| \leq C|x|^{k}$


For $I \subset Dom(f)$ open containing $0$, and $C > 0$.
 A: You do things very complicate...
1) Let $C>0$ such that $|f(x)|\leq C|x^k|$. In particular $$\left|\frac{f(x)}{x^{k-1}}\right|\leq C|x|\underset{x\to 0}{\longrightarrow }0$$
what prove the claim.
2) Let $\varepsilon>0$ and $\delta>0$ such that
$$\left|\frac{g(x)}{x}\right|<\varepsilon$$
if $|x|<\delta$.
In particular
$$\left|\frac{g(x)-g(0)}{x-0}-0\right|=\left|\frac{g(x)}{x}\right|<\varepsilon$$
if $|x|<\delta$, what prove the claim. Suppose $g$ is $m$ times differentiable. You can observe that $g'(0)=0$ and that 
$$0=\lim_{x\to 0}\left|\frac{g(x)}{x^m}\right|=\lim_{x\to 0}\frac{1}{m}\left|\frac{g'(x)}{x^{m-1}}\right|$$ by l'Hopital rule and the fact that $x\mapsto |x|$ is continuous at $x=0$. Therefore $g'(x)=o(x^{m-1})$ and thus, by induction $g^{(k)}(x)=0$ for all $k=0,...,m$.
3) By 2) we have that $h^{(n)}(0)=0$. Moreover, by continuity of $h^{(n+1)}$, $$\lim_{x\to 0}h^{(n+1)}(x)=\lim_{x\to 0}\frac{h^{(n)}(x)}{x}=h^{(n+1)}(0).$$ 
Then there is a $\delta>0$ such that
$$|h^{(n+1)}(x)|\leq 1+|h^{(n+1)}(0)|$$
if $|x|<\delta$. We can conclude that $h^{(n+1)}(x)=\mathcal O(1)$ and thus that $h(x)=\mathcal O(x^{n+1)}$. Indeed, it's very easy to prove that 
$$|h^{(n+1)}(x)|\leq C\implies |h^{(n)}(x)|\leq C(|x|+\underbrace{h^{(n)}(0)}_{=0})=C|x|\implies ...\implies |h(x)|\leq C|x|^{n+1}$$
if $|x|<\tilde \delta$ for a certain $\tilde\delta>0$.
