For every $h \in \mathbb R^n$ the limit $\lim_{t \rightarrow 0} \frac 1 t \big(f(x_0 + th)-f(x_0)\big)$ exists and is equal to $D_{f_{x_0}}(h)$? Suppose $U \subset \mathbb R^n$ is open and $f: U \rightarrow \mathbb R^m$ is differentiable in $x_0 \in U$.
I want to show that for every $h \in \mathbb R^n$ the limit $\lim_{t \rightarrow 0} \frac 1 t \big(f(x_0 + th)-f(x_0)\big)$ exists and is equal to $D_{f_{x_0}}(h)$.
I've made the following considerations:
I can write $h = \sum a_i e_i$. Then $$D_{f_{x_0}}(h) = D_{f_{x_0}}\left(\sum a_i e_i\right) = \sum a_i D_{f_{x_0}}(e_i) = \sum a_i \left(\frac {\partial f_i} {\partial x_1}, \ldots, \frac {\partial f_i} {\partial x_n}\right)^T.$$
But how can I see that the limit $\lim_{t \rightarrow 0} \frac 1 t \big(f(x_0 + th)-f(x_0)\big)$ exists and equals the above linear transformation? It would be easier if $f$ were a linear function.
 A: What follows is for $m=1$. For arbitrary $m$ the argument is componentwise.
Denote $L$ the differential of $f$ at $x_0$. That the differential does exist means by definition that the following limit exists:
$$
\lim_{u\to0}\frac{f(x_0+u)-f(x_0)-L(u)}{\|u\|}=0.
$$
Now fix any $h\ne0$ and for $t\to0$ we get $u=th\to0$. From the above limit we get
$$
0=\lim_{t\to0}\frac{f(x_0+th)-f(x_0)-L(th)}{\|th\|}=\lim_{t\to0}\frac{f(x_0+th)-f(x_0)-tL(h)}{|t|\|h\|}.
$$
Equivalently:
$$
0=\frac{1}{\|h\|}\lim_{t\to0}\Big|\frac{f(x_0+th)-f(x_0)-tL(h)}{t}\Big|=\frac{1}{\|h\|}
\lim_{t\to0}\Big|\frac{f(x_0+th)-f(x_0)}{t}-L(h)\Big|.
$$
Since $\|h\|\ne0$ we conclude
$$
\lim_{t\to0}\frac{f(x_0+th)-f(x_0)}{t}=L(h),
$$
as wanted.
Note: the other computations you suggest in fact help to see that in case directional derivatives define a linear form $L$, then that linear form is escalar product with the gradient. But this doesn't yet means the function is differentiable: this only comes from the limit stated at the beginning.
