If $f$ is uniformly differentiable $\Longrightarrow$ $f'$ is continuous

I'm not sure how to go about proving this theorem:

Let $U\subset \mathbb{R}^m$ (open set) and $f:U\longrightarrow \mathbb{R}^n$ a differentiable function such that:

$\forall \epsilon>0\,,\exists \delta>0:|\!|h|\!|<\delta,[x,x+h]\subset U \Longrightarrow |\!|f(x+h)-f(x)-f'(x)(h)|\!|<\epsilon |\!|h|\!|$

then it holds that $f':U\longrightarrow \mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ is continuous.We can also say that $f'$ is uniformly continuous?

Any hints would be appreciated.

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If you interchange the roles of $x$ and $x+h$ (and replace $h$ by $-h$), then you see that $$|f(x)-f(x+h)+f'(x+h)(h)|<\epsilon|h|,$$ which, upon combining with what you have, and using the triangle inequality, shows that $f'(x)$ and $f'(x+h)$ are $\epsilon$-close.

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Ok timur ,$\large||[f'(x+h)-f'(x)](\frac{h}{||h||})||<\epsilon$ ,but as you get to that $\large ||f'(x+h)-f'(x)||<\epsilon$ ? – felipeuni Sep 21 '12 at 18:52
@felipeuni: A similar manipulation involving 4 points $x$, $x+h$, $x+k$, and $x+h+k$ seems to work. – timur Sep 23 '12 at 1:15