determinant differentiable at identity I would like to prove that the determinant $det:\mathbb{M_n}\rightarrow\mathbb{R}$ is differentiable at the identity matrix with $(Ddet(I))(H)=tr(H)$. Using the definition of differentiability this boils down to showing that:
$$\displaystyle{\lim_{H \to 0}}\frac{\det(I+H)-det(I)-tr(H)}{||H||}=0 \space \space (\bigstar)$$
Now I'm aware of the formula: $\det(I+H)=1+tr(H)+\frac{(tr(H)^2-tr(H^2))}{2!}+\frac{(tr(H)^3-3tr(H)tr(H^2)+2tr(H^3))}{3!}\dots$
but I don't see any way to show that: $$\displaystyle{\lim_{H \to 0}}\frac{\frac{(tr(H)^2-tr(H^2))}{2!}+\frac{(tr(H)^3-3tr(H)tr(H^2)+2tr(H^3))}{3!}+\dots}{||H||}=0$$

$1)$ I'm aware of the directional derivative aproach (see here for example link) but that can be used only after we know that the derivative exists, to show that $(Ddet(I))(H)=tr(H)$.
$2)$ I'm also aware of the proof treating the determinant as a polynomial function of $n^2$ variables.

But I would like to know if there exists a proof using just the definition of derivative $(\bigstar)$.
Thank you!
 A: $\DeclareMathOperator{\tr}{tr}$
Assume $\|A\| = \|A\|_F$, i.e. we endow $M_{n}(\mathbb{R})$ with the Frobenious norm, then it follows
\begin{align}
\|H\|_F = (\lambda_1^2+\cdots+\lambda_n^2)^{1/2}
\end{align}
where $\lambda_i$ are the eigenvalues of $H$. Then we see that
\begin{align}
\det(I+H) = \det( I+\varepsilon \hat H)
\end{align}
where $\varepsilon = \|H\|_F$ and $\hat H = H/\varepsilon$. Note that
\begin{align}
\det(I+\varepsilon \hat H) = 1+\varepsilon \tr\hat H +\mathcal{O}(\varepsilon^2) 
\end{align}
as shown here. Hence it follows
\begin{align}
\left|\frac{\det(I+H)-1-\tr H}{\varepsilon}\right|=\left|\frac{\det(I+H)-1-\varepsilon \tr \hat H}{\varepsilon}\right| \leq  M\varepsilon.
\end{align}
Note that $M$ is a uniform bound on the sphere $\|A\|=1$. 
To be even more specific, we see that
\begin{align}
\frac{\tr(H)^2-\tr(H^2)}{2!} = \frac{\|H\|^2}{2!}\left(\tr(\hat H)^2-\tr(\hat H^2) \right)
\end{align}
where
\begin{align}
\left|\tr(\hat H)^2-\tr(\hat H^2) \right| \leq |\tr(\hat H)|^2+|\tr(\hat H^2)| \leq C_n2
\end{align}
and 
\begin{align}
|\tr(\hat H)^3-3\tr(\hat H)\tr(\hat H^2)+2\tr(\hat H^3)| \leq C_n6=C_n3!.
\end{align}
where $C_n$ is a constant that only depends on the dimension. 
Likewise, for the other terms.
A: $\DeclareMathOperator\tr{tr}$Let $\lambda_M$ be the largest magnitude of the eigenvalues of $H$.
Let $\sigma_M$ be the largest singular value of $H$, which is the square root of the largest eigenvalue of $H^TH$.
Then $\|H\|=\sigma_M \ge \lambda_M$, $|\tr(H)|\le n\lambda_M\le n\sigma_M$, and $\tr(H^2) \le n\lambda_M^2\le n\sigma_M^2$.
It follows that:
$$\lim_{H\to 0} \frac{(\tr H)^2}{\|H\|} = \lim_{\sigma_M\to 0} \frac{(\tr H)^2}{\sigma_M}
\le \lim_{\sigma_M\to 0} \frac{(n\sigma_M)^2}{\sigma_M} =0$$
and:
$$\lim_{H\to 0} \frac{\tr(H^2)}{\|H\|} = \lim_{\sigma_M\to 0} \frac{\tr(H^2)}{\sigma_M}
\le \lim_{\sigma_M\to 0} \frac{n\sigma_M^2}{\sigma_M} =0$$
The same applies to all higher powers, and your limit is therefore $0$.
