Confusion with the Taylor theorem for vetor-valued multivariable functions

Question

Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a smooth function.

If I pick a point $a \in \mathbb{R}^n$, then I think I can apply the Taylor theorem to $f$ by \begin{equation} f(x)=f(a)+[Df(a)] \cdot (x-a) + \text{remainder } R_1(x,a) \end{equation} where $Df(a)$ is the Jacobian matrix of $f$ at $a$ and $[Df(a)] \cdot (x-a)$ means that $ n\times n$ matrix is multiplied to $n \times 1$ column vector. Also, $R_1(x,a)$ is $\mathbb{R}^n$-valued smooth function of $x$ for the given $a$,

Now, if I write like this, I wonder if \begin{equation} \frac{\lVert R_1(x,a) \rVert}{\lVert x-a \rVert^2} \end{equation} is bounded as $x \to a$.

Generalizing from one dimensions, I guess this must be true, but I am extremely confused due to many variables interwined..

Could anyone please clarify?

If you assume $f$ is smooth (weaker conditions suffice) then sure it is true. There are several statements for Taylor’s theorem in the Banach-space setting, and bounds on the remainder, see eg Bound on remainder for vector valued Taylor series. For a more detailed explanation, I suggest reading Henri Cartan’s text. — peek-a-boo, May 25 '23 at 00:11
Generally the remainder is $o(| x-a |)$, not $o(| x-a |^2)$. — copper.hat, May 25 '23 at 00:33

Confusion with the Taylor theorem for vetor-valued multivariable functions

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