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In the textbook "Differential Topology" I came across the following statement:

Suppose that $f$ is a smooth map of an open set in $\mathbb{R}^n$ into $\mathbb{R}^m$ and $x$ is any point in its domain. Then for any vector $h \in \mathbb{R}^n,$ the derivative of $f$ in the direction $h,$ taken at the point $x,$ is defined by the conventional limit

$$df_x(h) = \lim_{t \to 0} \frac{f(x+th)-f(x)}{t}$$

With $x$ fixed, we define a mapping $df_x: \mathbb{R}^n \to > \mathbb{R}^m$ by assigning to each vector $h \in \mathbb{R}^n$ the directional derivative $df_x(h) \in \mathbb{R}^m.$ Note that this map, which we call the derivative of $f$ at $x,$ is defined on all of $\mathbb{R}^n$ even though $f$ need not be.

Why is the last statement true? How can the derivative of $f$ at $x$ exist, or the above limit, if $f(x)$ is not defined?

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    What it is saying is that the map $df_x$ itself is defined on all of $\Bbb R^n$, though $f$ might not be defined on all of $\Bbb R^n$. – 2017-01-27

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The copy of $\mathbb{R}^{n}$ that the derivative, $\text{d}f_{x}$ is acting on is not the copy of $\mathbb{R}^{n}$ that the function $f$ itself is working on.

To compute the value $\text{d}f_{x}(h) \in \mathbb{R}^{m}$ we need to know the value of $f$ in an open neighborhood of the point $x$, independently of the magnitude of the vector $h$, this is visible from the formula you wrote down.