about the differentiability : the general case

Let $U$ be an open set in $\mathbb{R}^{n}$ and $f :U \subset \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a given function. We say that $f$ is differentiable at $x_{0}\in U$ if the partial derivatives of $f$ existi at $x_{0}$ and if $$\displaystyle \lim_{x \rightarrow x_{0}} \frac{\|f(x)-f(x_{0})-T(x-x_{0})\|}{\|x-x_{0}\|}=0$$ where $T=Df(x_{0})$ is the $m \times n$ matrix with elements $\displaystyle \frac{\partial f_{i}}{\partial x_{j}}$ evaluated at $x_{0}$ and the $T(x-x_{0})$ means the product of $T$ with $x-x_{0}$ (regarded as a column matrix). We call $T$ the derivative of $f$ at $x_{0}$.

Now I consider a particular case($m=n=2$) $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$

Following the definition I obtain :

$$\displaystyle f(a,b)-f(a_{0}, b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} & \frac{\partial f_{1}}{\partial b}\\ \frac{\partial f_{2}}{\partial a} & \frac{\partial f_{2}}{\partial b} \end{pmatrix} \begin{pmatrix}a-a_{0} \\ b-b_{0} \end{pmatrix} = f(a,b)-f(a_{0},b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} \cdot(a-a_{0})+\frac{\partial f_{1}}{\partial b}\cdot(b-b_{0})\\ \frac{\partial f_{2}}{\partial a}\cdot(a-a_{0})+\frac{\partial f_{2}}{\partial b}\cdot(b-b_{0}) \end{pmatrix}$$ where $f(a,b)=(f_{1}(a,b),f_{2}(a,b))$.

My question is : How can I compute this limit because the last element is a matrix and first two aren't. And why $f(x)-f(x_{0})-T(x-x_{0})$ I have to put into $\| \|$ ? I have an idea why I have to put into norm but I'm not sure and can you give an concrete example how I compute the limit - for example $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2, f(a,b)=(a^2+b^2, a^2+b^2)$ when $(a,b) \rightarrow (1,1)$.

Thanks :)

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The elements $f(a,b)$ and $f(a_0,b_0)$ are already vectors. – emiliocba Aug 25 '12 at 10:09
All three summands are $2\times 1$-matrices, that is, vectors in $\mathbb R^2$. So it makes sense to add them. – Rasmus Aug 25 '12 at 10:10
@emiliocba yes, I know the element $f(a,b)$ and $F(a_{0},b_{0})$ are vector, but I must see these two vector like column vectors ? – Iuli Aug 25 '12 at 10:13
The definition of differentiable functions is not correct. The mistake is in the hypothesis "if the partial derivatives of f existi at $x_0$". See en.wikipedia.org/wiki/Differentiable_function – emiliocba Aug 25 '12 at 10:13
@emiliocba I think the definition is correct. How can you have matrix $T$ if you haven't partial derivatives? – Iuli Aug 25 '12 at 10:20

Let me compute for the example, $f(x,y)=(x^2+y^2, x+y)$. We write $f_1(x,y)=x^2+y^2$ and $f_2(x,y)=x+y$. Then $$\frac{\partial f_1}{\partial x}(x,y) = 2x,\quad \frac{\partial f_1}{\partial y}(x,y) = 2y,\quad \frac{\partial f_2}{\partial x}(x,y) = 1,\quad \frac{\partial f_2}{\partial y}(x,y) = 1.$$ Since these four functions are continuous then $f$ is differentiable. THEN $T$ is already the Jacobian matrix: $$T=\begin{pmatrix} 2x& 2y\\ 1&1\end{pmatrix}.$$ Finally $$f(x,y)-f(x_0,y_0)-T\begin{pmatrix}x-x_0\\y-y_0\end{pmatrix}= \begin{pmatrix}x^2+y^2-x_0^2-y_0^2-2x_0(x-x_0)-2y_0(y-y_0) \\ x+y-x_0-y_0 -(x-x_0+y-y_0)\end{pmatrix},$$ taking limits $$\lim_{(x,y)\to(x_0,y_0)} \frac{(x-x_0)^2+(y-y_0)^2}{\sqrt{(x-x_0)^2+(y-y_0)^2}}= \lim_{(x,y)\to(x_0,y_0)} \sqrt{(x-x_0)^2+(y-y_0)^2}=0.$$