determinant differentiable at identity

$\DeclareMathOperator\tr{tr}$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 $0$.

$\DeclareMathOperator{\tr}{tr}$ 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.