Why is the total differential divided by the norm of h bounded? I'm trying to proof the product rule for functions $f: R^{n} \to R$. There is already a good thread on this -  total differential of $f+g$, $fg$ and $\frac fg$.
However, I am not sure about the last step, i.e. showing that
$$\lim \limits_{h \to 0} \frac{(f(x+h)-f(x))dg_{x}(h)}{\lVert h \rVert}=0$$
As the other thread mentions in the comments
\begin{align} & \lim \limits_{h \to 0} \frac{(f(x+h)-f(x))dg_{x}(h)}{\lVert h \rVert} \\
& =\lim \limits_{h \to 0} (f(x+h)-f(x)) \lim \limits_{h \to 0} \frac{dg_{x}(h)}{\lVert h \rVert} \\
&
\end{align}
and the first limit converges to $0$ by continuity of $f$ (since f is totally differentiable). So it remains to show that the second term is bounded.
But I am not sure how to do this.

I have given this another thought and have come up with a solution.
Since $dg_{x}(h)$ is a linear map on a finite-dimensional vector space, it is a bounded linear operator which means that $\lvert dg_{x}(h)\rvert<C \|h\|, C \in \mathbb{R}$ for all $h$ (see https://en.wikipedia.org/wiki/Discontinuous_linear_map).
Therefore, $\frac{dg_{x}(h)}{\|h\|} \leq \frac{\lvert dg_{x}(h)\rvert}{\|h\|}<C$.
