Suppose $f$ is differentiable in $\overline{x}\in D$ then exists $M>0$ and $r>0$ such that... Let $D\subset \mathbb{R}^n$ a open set and let $f:\,D\rightarrow\mathbb{R^{\text{m}}}$a function. Suppose $f$ is differentiable in $\overline{x}\in D$ then exists $M>0$ and $r>0$ such that
$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})||}{||\overline{h}||}\leq M$
If $0<||\overline{h||}<r$
My work:

$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})||}{||\overline{h}||}=\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})-df_{x_{0}}(\overline{h})|}{||\overline{h}||}\leq\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})||}{||\overline{h}||}+\frac{||df_{x_{0}}(\overline{h})||}{||\overline{h}||}$

As $df_{x_{0}}$ is a linear transformation, then exists $M_{1}>0$ such that >$df_{x_{0}}(\overline{h})\leq M_{1}$. 

Moreover, $df_{x_{0}}(\overline{h})\leq M_{1}\times||\overline{h}||$
Then exists $K_{1}>0$ such that
$\frac{||df_{x_{0}}(\overline{h})||}{||\overline{h}||}<K_{1}$
for all $\overline{h}\in\mathbb{R^{\text{n}}}\,$if$||\overline{h}||\neq\overline{0}$
For hypothesis $f$ is differentiable in $\overline{x}$ then we have
$lim_{\overline{h}\rightarrow\overline{0}}\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})||}{||\overline{h}||}=0$
Then exists $K_{2}>0$ such that 
$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})||}{||\overline{h}||}<K$
If $M=K_{1}+K_{2}$ then we have
$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})||}{||\overline{h}||}=\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})-df_{x_{0}}(\overline{h})|}{||\overline{h}||}\leq\frac{||f(\overline{x}+\overline{h})-f(\overline{x})|+df_{x_{0}}(\overline{h})||}{||\overline{h}||}+\frac{||df_{x_{0}}(\overline{h})||}{||\overline{h}||}=K_{1}+K_{2}=M$
In consequence:
$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})||}{||\overline{h}||}<M$
I have two problems with this proof

1) That $r>0$ is for what? I can not understand its importance in this exercise... What happen if $||\overline{h||}>r$ ?
2) My teacher give me this hint. "As $df_{x_{0}}$ is a linear transformation, then exists $M_{1}>0$ such that >$df_{x_{0}}(\overline{h})\leq M_{1}$"
Then, my question is. if $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear transformation, always can i found a $M\in \mathbb{R}$ such that $T(x)<M$ for all $x\in \mathbb{R}$

 A: 1) The $r>0$ means something locally, in general we don't have such nice thing $\dfrac{|f(x+h)-f(x)|}{|h|}\leq M$ globally for $|h|$, which means that for large $|h|$ this inequality would fail.
2) Because you have restricted locally, and for bounded linear operator $T$, one has $|Tx|\leq M|x|$.
3) Not for all $x\in{\bf{R}}$, take for example $T(x)=x$, saying that $|T(x)|<M$ for all $x$ means that $|x|<M$ for $x\in{\bf{R}}$, this is clearly not true. As what has been said in 2), $|Tx|=|x|\leq M|x|$ is true, here $M=1$.
A: The main step is to demonstrate the equivalent of MVT in higher dimensions, that is:
$||f(\overline{x}+\overline{h})-f(\overline{x})||\leq||df_{x_{0}}\overline{h}||$
Since $df_{x_{0}}$ is a linear transformation (i.e. a matrix) then $\exists M$ s.t. $||df_{x_{0}}\overline{h}||\leq M||\overline{h}||$.
Thus, it follows immediately that:
$$\frac{||f(\overline{x}+\overline{h})-f(\overline{x})||}{||\overline{h}||}\leq M$$
NOTE
I suppose that D should be assumed to be convex (i.e. path-connected).
OTHER DISCUSSION on MVT
A generalization of the mean value theorem?
