Swapping limit with inner product. The exercise is:

Given $\xi : U \subset \Bbb R^m \rightarrow \Bbb R$, $U$ open, given by $\xi (x) = \langle f(x), g(x) \rangle$, where $f,g: U \rightarrow \Bbb R^p$ are differentiable functions, compute $\mathrm{d}\xi_x(h)$, for an arbitrary $h \in \Bbb R^m$. Here, $\langle \cdot, \cdot \rangle$ denotes the inner product. 

The exercise also asks to use the definition. I'll put my work here, also. $$\begin{align} \mathrm{d}\xi_x(h) &= \lim_{t \to 0} \frac{\xi(x + th) - \xi(x)}{t} \\ &= \lim_{t \to 0} \frac{\langle f(x + th), g(x + th)\rangle - \langle f(x), g(x) \rangle}{t} \\ &= \lim_{t \to 0} \frac{\langle f(x + th), g(x + th)\rangle - \langle f(x), g(x + th)\rangle + \langle f(x), g(x + th) \rangle - \langle f(x), g(x) \rangle}{t} \\ &= \lim_{t \to 0} \left\langle \frac{f(x + th) - f(x)}{t}, g(x + th) \right\rangle + \left\langle f(x), \frac{g(x + th) - g(x)}{t} \right\rangle \end{align}$$
There's a great resemblance to the usual product rule for derivatives. If I indeed can swap the limit with the inner product, we would get nicely $$\mathrm{d}\xi_x(h) = \langle \mathrm{d}f_x(h), g(x) \rangle + \langle f(x), \mathrm{d}g_x(h) \rangle $$
But how can I justify this accordingly? This exercise is from page $4$ of an Analysis book, so I'm not sure how many well-known results we can use. Perhaps use the results from limits of composition of functions, and continuity of the inner product somehow? Can someone please explain it to me? Thank you (:
 A: Since you're in a finite dimensional space you could look at the components i.e.
$$
\begin{eqnarray}
&&
\lim_{t \to 0} \left\langle \frac{f(x + th) - f(x)}{t}, g(x + th) \right\rangle \\
& = & \lim_{t \to 0} \sum_{i=1}^n \frac{f(x + th)_i - f(x)_i}{t}g(x+th)_i \\
& = & \sum_{i = 0}^n \lim_{t \to 0} \frac{f(x + th)_i - f(x)_i}{t}g(x+th)_i \\
& = & \sum_{t = 0}^n \nabla_h f(x)_i g(x)_i \\
& = & \left\langle \nabla_h f (x) , g(x) \right\rangle
\end{eqnarray}
$$
and
$$
\begin{eqnarray}
&& \lim_{t \to 0} \left\langle f(x), \frac{g(x + th) - g(x)}{t} \right\rangle \\
& = & \lim_{t \to 0} \sum_{i=1}^n f(x)_i \frac{g(x + th)_i - g(x)_i}{t} \\
& = & \sum_{i=1}^n f(x)_i \lim_{t \to 0} \frac{g(x + th)_i - g(x)_i}{t} \\
& = & \sum_{i=1}^n f(x)_i \nabla_h g(x)_i \\
& = & \left\langle f(x), \nabla_h g(x) \right\rangle
\end{eqnarray}
$$
So now we're able to write that
$$
\begin{eqnarray}
&&
\lim_{t \to 0} \left\langle \frac{f(x + th) - f(x)}{t}, g(x + th) \right\rangle + \left\langle f(x), \frac{g(x + th) - g(x)}{t} \right\rangle \\
& = & \lim_{t \to 0} \left\langle \frac{f(x + th) - f(x)}{t}, g(x + th) \right\rangle + \lim_{t \to 0} \left\langle f(x), \frac{g(x + th) - g(x)}{t} \right\rangle \\
& = & \left\langle \nabla_h f (x) , g(x) \right\rangle + \left\langle f(x), \nabla_h g(x) \right\rangle
\end{eqnarray}
$$
