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I'm trying to understand a specific proof for the notion that differentiability implies continuity in multivariate space.

After having defined differentiability as:

Let $U \subset \mathbb{R}^n$ be open, $\vec f:U \rightarrow \mathbb{R}^m$ be a mapping. Now we call $\vec f$ differentiable in $\vec x_0 \in U$, if there exists a linear mapping $A: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that:

$\vec f(\vec x_0 + \vec h) - \vec f(\vec x_0) - A \vec h = \vec \phi (\vec h)$ with $\vec \phi (\vec h) = 0 (||\vec h||) \Leftrightarrow \lim_{\vec h \rightarrow \vec 0} \frac{||\vec \phi (\vec h)||}{||\vec h||} = 0$, which leads us to the derivative in the form: $A\vec h = d \vec f(\vec x_0) \vec h$. (Which I think I understand but am not 100% sure.)

Now for the proof of differentiability implies continuity of $\vec f: U \subset \mathbb{R}^n \rightarrow \mathbb{R}^m$ in $\vec x_0 \in U$, we have:

$lim_{x \rightarrow x_0} \vec f (\vec x) = \vec f (\vec x_0) + lim_{x \rightarrow x_0} A(\vec x - \vec x_0) + lim_{x \rightarrow x_0} \vec \phi (\vec x - \vec x_0)$, which, according to my source, yields $\vec f(\vec x_0) + \vec 0 + \vec 0$.

Now, I understand the implication of the proof, and also that the error term goes to $0$, as that is by definition, but I just don't understand how the A function will become zero. Maybe I'm missing something from it's definition as well? Could someone perhaps explain the definition or maybe intuition behind it, any further? Thanks for any answer.

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Hint:

As $A$ is a linear operator from one finite-dimensional space to another there exists a constant $C$ such that $\|A(x - x_0)\| \leqslant C \|x - x_0\|$.

This is easy to prove by expanding with respect to the standard basis vectors for $\mathbb{R}^n$ and $\mathbb{R}^m.$

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