Approximation of $f'(x)$ using Taylor Let $I \subseteq \mathbb{R}$ open interval and $f: I \to \mathbb{R}$ twice differentiable with continuous second derivative in a open neighborhood of $x \in I$. Show that there are positive constants $\delta$ and $c$ such that
$$\left\lvert f'(x) - \dfrac{f(x+h)-f(x)}{h}\right\rvert\le ch,$$
for every $0 \lt h \lt \delta$. Show that constant $c$ can be chosen independently from $h$.
My take: I tried to approximate $f(x+h)$ and $f(x-h)$ by Taylor approximation and then sum the terms or take their difference, but I wasn't able to find the desired result. Any thoughts? Thanks!
 A: Let $x \in I$ and $U_{x}$ be an open neighborhood of $x$ such that $f$ is twice differentiable with continuous derivatives in $U_{x}$. Since $U_{x}$ is a non-empty open set, $\exists \delta \gt 0:(x-\delta,x+\delta) \subseteq U_{x}$ and $f$ is twice differentiable on $[x-\delta,x+\delta]$.
Let $h \in \mathbb{R}:0 \lt h \lt \delta$. Notice that $f$ is differentiable in $[x-\delta,x+\delta]$ with $f'$ continuous in this interval and that $f''$ exists in $(x-\delta,x+\delta)$.
By Taylor's Theorem, $\exists \xi \in (x,x+h)$ such that:
$f(x+h) = f(x) + f'(x)h + \frac{f''(\xi)h^{2}}{2}$
$\frac{f(x+h)-f(x)}{h}=f'(x)+\frac{f''(\xi)h}{2}$
$\bigg|f'(x) - \frac{f(x+h)-f(x)}{h}\bigg|=\bigg|f''(\xi)\bigg|\frac{h}{2}$
Since $f''$ is continuous in $[x-\delta,x+\delta]$, which is a compact set, its image is also compact, therefore, limited. Hence, $\exists c \gt 0:|f''(y)| \leq 2c, \forall y \in [x-\delta,x+\delta]$. Since $(x,x+h) \subseteq [x-\delta,x+\delta]$, we have:
$\bigg| f'(x)-\frac{f(x+h)-f(x)}{h} \bigg| = |f''(\xi)|\frac{h}{2} \leq 2c\frac{h}{2}=ch$
A: The inequality as stated is not true. For instance let $f(t)=t$, $x=1$. Then
$$
f'(x)-\frac{f(x-h)-f(x)}h=1-\frac{x-h-x}h=2.
$$
It does work with a plus sign, though. "Taylor" here is simply the Mean Value Theorem. If you let 
$$
g(h)=\begin{cases} \frac{f(x+h)-f(x)}h,&\ h\ne0\\ \ \\ f'(x),&\ h=0\end{cases}
$$
The hypotheses guarantee that $g$ is differentiable on $I$. By the Mean value Theorem, 
$$
|g(h)-g(0)|=|g'(y)\,(h-0)|=|g'(y)|\,h
$$
for some $y$ between $0$ and $h$. If we choose any $\delta>0$ such that $[x-\delta,x+\delta]\subset I$, then $g'$ is bounded (continuous on a compact set) so there exists $c$ with $|g'(y)|\leq c$ for all $y\in [x-\delta,x+\delta]$. Thus
$$
\left|f'(x)-\frac{f(x+h)-f(x)}{h}\right|=|g(0)-g(h)|\leq ch. 
$$
A: This is an easy consequence of Mean value Theorem. Write $f(x+h)-f(x)$ as $hf'(z)$ for some $z$ between $x$ and $x+h$. Then $|f'(x)-\frac {f(x+h)-f(x)} h =|f'(x)-f'(z)|$. Apply Mean Value Theorem again. You get $|x-z||f''(u)|$ for some $u$ between $x$ and $z$. Just use the fact that $f''$ is bounded in some interval around $x$ and $|x-z|\leq |h|$. 
