Question on Infinitely Differentiable Functions Let $f:(-1,1)\rightarrow\mathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:
(1) $\vert f^{(n)}(x)\vert \le n!$ for every $x\in (-1, 1)$ and for every $n\in \mathbb N$
(2) $f'(\frac{1}{m+1})=0$ for every $m\in \mathbb N$
I need to:
Find the value of $f^{(n)}(0)$ for each $n\in \mathbb N$
Determine the value of $f(x)$ for every $x\in(-1,1)$
I claim that $f^{(n)}(0)=0$ for all $n\in \mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $m\rightarrow \infty$. However, I am not sure on how to prove the inductive step.
I note that from (1), we have $f(x)\le\frac{1}{1-x}$ by Taylor's Theorem.
Any help is greatly appreciated!
 A: Rolle's theorem implies that $f''$ has a zero  in the interval $(\frac{1}{m+2}, \frac{1}{m+1})$
for each $m \in \Bbb N$, therefore $f''(0) = 0$.
The same argument can be repeated to show that each derivative $f^{(k)}$ has
a sequence of zeros converging to $0$.
(The estimate $\vert f^{(n)}(x)\vert \le n!$ is not needed for this
conclusion.)
A: Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)
For any $m\in\mathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_m\in (-1,1)$ such that 
$$f'\left(\frac{1}{m+1}\right)=f'(0)+f''(0)\cdot \frac{1}{m+1}+\frac{f'''(t_m)}{2}\cdot \frac{1}{(m+1)^2}.$$
Since $f'\left(\frac{1}{m+1}\right)=0$ and we already know that $f'(0)=0$, it follows
$$f''(0)=-\frac{f'''(t_m)}{2}\cdot \frac{1}{(m+1)}$$
which implies that as $m\to+\infty$,
$$|f''(0)|=\frac{|f'''(t_m)|}{2}\cdot \frac{1} {(m+1)}\leq\frac{M_3} {2(m+1)}\to 0.$$
where $M_3=\max_{x\in[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption  $\vert f^{(n)}(x)\vert \le n!$ in $(-1,1)$ is not needed). 
