Theorem 9.17 Baby Rudin 
9.17 Theorem Suppose f maps an open set $⊂^{n}$ into $^{m}$, and $f$ is differentiable at a point $∈$. Then the partial derivatives $_{j}_{i}()$ exist, and
  $$\begin{align}
\mathbf{f'(x)e_j}=\sum\limits_{i=1}^m(D_jf_i)\mathbf{(x)u_i} &&(1\le j\le n)\tag{27}
\end{align}$$
$_{i}$′s and $_{j}$′s are standard bases of $^{n}$ and $^{m}$ respectively.
Proof
  Fix . Since $f$ is differentiable at $$,
  $$\mathbf{f(x+}t\mathbf{e_j)}-\mathbf{f(x)}=\mathbf{f'(x)(}t\mathbf{e_j)}+\mathbf{r(}t\mathbf{e_j)}\tag{28}$$
  where $∣(_{j})∣/→0$ as $→0$. The linearity of $′()$ shows therefore that
  $$\lim_{t\to 0}\sum\limits_{i=1}^m\frac{f_i(\mathbf{x}+t\mathbf{e_j})-f_i(\mathbf{x})}{t}\mathbf{u_i}=\mathbf{f'(x)e_j.}\tag{29}$$
  It follows that each quotient in this sum has a limit, as $→0$(see Theorem 4.10), so that each $(_{j}_{i})()$ exists, and then (27) follows from (29).

Here is 4.10Theorem
4.10 and 4.14 Theorem in Rudin
My question is how we exactly use theorem 4.10 to get $(29)$? I have looked other related questions but they just mentioned this allows us to interchange limit and sum, I want to know the detail.
Thanks in advance
 A: The calculation in a specific case will show directly where the theorem is used. Let's say $f:\mathbb R^2\to \mathbb R^2$ and take $j=1.$  Then, 
$\lim_{t\to 0}\sum\limits_{i=1}^2\frac{f_i(\mathbf{x}+t\mathbf{e_1})-f_i(\mathbf{x})}{t}\mathbf{u_i}=\lim_{t\to 0}\left(\frac{f_1(\mathbf{x}+t\mathbf{e_1})-f_1(\mathbf{x})}{t}\mathbf{u_1}+\frac{f_2(\mathbf{x}+t\mathbf{e_1})-f_2(\mathbf{x})}{t}\mathbf{u_2}\right)=\lim_{t\to 0}\left(\frac{f_1(\mathbf{x}+t\mathbf{e_1})-f_1(\mathbf{x})}{t}(1,0)+\frac{f_2(\mathbf{x}+t\mathbf{e_1})-f_2(\mathbf{x})}{t}(0,1)\right)=\lim_{t\to 0}\left(\frac{f_1(\mathbf{x}+t\mathbf{e_1})-f_1(\mathbf{x})}{t},\frac{f_2(\mathbf{x}+t\mathbf{e_1})-f_2(\mathbf{x})}{t}\right)$
It is right here that Theorem $(4.10)$ in Rudin is used, to pass the limit through to the coordinates. You get 
$\begin{pmatrix}
(f_1)_x(\bf x) &(f_2)_x(\bf x) 
\end{pmatrix}$.
Now, to verify the formula, we calculate 
$f'(\bf x)e_1=\begin{pmatrix}
(f_1)_x(\bf x) &(f_1)_y(\bf x) \\ 
(f_2)_x(\bf x) & (f_2)_y(\bf x)
\end{pmatrix}\begin{pmatrix}
1\\0 \end{pmatrix}=\begin{pmatrix}
(f_1)_x(\bf x) &(f_2)_x(\bf x) 
\end{pmatrix}$, as desired. 
