Questions on Strongly Differentiability. 
Definition: Let $U\subseteq \Bbb R^m$ be an open set. Let $f: U \to \Bbb R^n$ be a function and $T: \Bbb R^m \to \Bbb R^n$ be a linear
  transformation. We say that $f$ is strongly differentiable at $x_0$,
  with derivative $T$, if for every $\epsilon>0$ there is $\delta >0$
  s.t for any $x_1, x_2$ in $ \delta$-nbd of $x_0$ i.e
  $\|x_i-x_0\|<\delta$ for $i=1,2$ imply
  $\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\epsilon\|x_2-x_1\|$.
Problem 1: Show that if $f$ is strongly differentiable at $x_0$ then it satisfies
  Lipschitz condition in a neighbourhood of $x_0$.
Problem 2 Show that if f is  differentiable in a neighbourhood $U$ of $x_0$ & if $Df$ is continuous as a function of $x$ at $x_0$ then $ f$ is strongly differentiable at $x_0$  .

My attempt is:
1) $\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\epsilon\|x_2-x_1\|\Rightarrow \|f(x_2)-f(x_1)\|<\epsilon\|x_2-x_1\| + \|T(x_2-x_1)\|$ .
Can anyone help me? What do I have to do next?
2) $||f(x_2)-f(x_1)-T(x_2-x_1)||<||f(x_2)-f(x_1)||+ ||T(x_2-x_1)||$ [because of definition of norm T, but why $||f(x_2)-f(x_1)||<\epsilon''||x_2-x_1||$]
Can anyone help me? What do I have to do next?
 A: I will answer only Problem 1, since Problem 2 was separately asked at Show that a function is strongly differentiable if it is continuously differentiable..
As PhoemeuX hinted, you should use the norm of $T$ (which is often written $\lVert{T}\rVert$ but which PhoemeuX wrote $L$); this exists because $\{x \mid \lVert{x}\rVert \leq 1\}$ is compact (and nonempty) and $T$ is continuous, which in turn are because $\mathbb{R}^m$ has finite dimension.  (And in fact, when people work in infinite-dimensional Banach spaces, they explicitly require $\lVert{T}\rVert$ to exist (as a finite supremum if not a maximum) when defining a strongly differentiable function between such spaces, and then the Lipschitz condition still follows.)
To make this answer complete, I'll copy the definition of $\lVert{T}\rVert$ from PhoemeuX's comment (with a tiny change from $\|x\| = 1$ to $\|x\| \leq 1$ to automatically handle the case $m = 0$): $\lVert{T}\rVert = \max_{\lVert{x}\rVert \leq 1} \lVert{Tx}\rVert$.  Then $\lVert{Ty}\rVert = \lVert{y}\rVert\,\lVert{T(y/\lVert{y}\rVert)}\rVert \leq \lVert{T}\rVert\,\lVert{y}\rVert$ (unless $y = 0$, when $\lVert{Ty}\rVert \leq \lVert{T}\rVert\,\lVert{y}\rVert$ still holds).
Then your argument can be completed:
$$\lVert{f(x_2) − f(x_1)}\rVert < \epsilon\,\lVert{x_2−x_1}\rVert + \lVert{T(x_2−x_1)}\rVert \leq \epsilon\,\lVert{x_2−x_1}\rVert + \lVert{T}\rVert\,\lVert{x_2−x_1}\rVert \leq (\epsilon+\lVert{T}\rVert)\,\lVert{x_2−x_1}\rVert$$
so pick your favorite $\epsilon > 0$, get the corresponding $\delta$-neighbourhood, and use $\epsilon+\lVert{T}\rVert$ as the Lipschitz constant.  (Thus you can make the Lipschitz constant as close to $\lVert{T}\rVert$ as you like by using a sufficiently small neighbourhood.  To see that you cannot always make it $\lVert{T}\rVert$ exactly, use $m,n = 1$, $f(x) = x^2$, $x_0 = 0$.)
