Equidifferentiability and Taylor series expansions

Question

Suppose the sequence of vector valued functions $\{ {\bf f}_n \}$ are equidifferentiable at ${\bf x}_0$. In other words:

$$\lim_{{\bf h} \to {\bf 0}} \max_n \frac{\left\Vert {\bf f}_n({\bf x}_0+{\bf h}) - {\bf f}_n({\bf x}_0) - \triangledown {\bf f}_n({\bf x}_0) {\bf h} \right\Vert}{\left\Vert {\bf h} \right\Vert} = 0$$

Suppose we have ${\bf x}_n$ such that ${\bf x_n} \to {\bf x}_0$ as $n \to \infty$.

Question. Is it true than we can then perform the following Taylor expansion

$$ {\bf f}_n({\bf x}_n) = {\bf f}_n({\bf x}_0) + \triangledown {\bf f}_n({\bf x}_0) ({\bf x}_n - {\bf x}_0) + o({\bf x}_n - {\bf x}_0)$$

Answer 1

It just follows from the rule $o(o(1)) = o(1)$. Denote

$$r_n({\bf h}) = \frac{\left\Vert {\bf f}_n({\bf x}_0+{\bf h}) - {\bf f}_n({\bf x}_0) - \triangledown {\bf f}_n({\bf x}_0) {\bf h} \right\Vert}{\left\Vert {\bf h} \right\Vert} $$

Then, we have, from equidifferentiability,

$$ \max_n r_n({\bf h}) = o({\bf h})$$

Hence, for any ${\bf h} \to 0$,

$$ {\bf f}_n({\bf x}_0 + {\bf h}) = {\bf f}_n({\bf x}_0) + \triangledown {\bf f}_n({\bf x}_0) {\bf h} + o({\bf h})$$

Replacing ${\bf h}$ by ${\bf x}_n -{\bf x}_0$ ,

$$ {\bf f}_n({\bf x}_n) = {\bf f}_n({\bf x}_0) + \triangledown {\bf f}_n({\bf x}_0) ({\bf x}_n - {\bf x}_0) + o({\bf x}_n - {\bf x}_0)$$