Suppose I have a matrix-valued, continuous function
$$A\colon [0,\infty) \to \mathbb R^{n\times n},\qquad h\mapsto A(h).$$
I know that for the limit $h\to 0$ the matrix is invertible:
$$\det\left(\lim_{h\to0}A(h)\right) \neq 0.$$
Does this imply that there is a $H>0$ and a $K>0$ with:
\begin{alignat}{99} \det\bigl(A(h)\bigr)&\neq 0,\qquad&\forall h&\in[0,H)\\ \|A(h)\| &\leq K,\qquad& \forall h&\in[0,H) \\ \|A^{-1}(h)\| &\leq K,\qquad& \forall h&\in[0,H) \end{alignat}
My thoughts are that that is true, because the determinant and the inversion are continuous functions. Am I right or is there something I have missed?
That $A(h)$ is invertible for small enough $h$ is clearly true, as you said, because $A$ and $\det$ are continuous.
Now, for the boundedness, I assume that the matrix norm you use, is the standard spectral norm.
Since $A$ is continuous, $B(h)=A(h)A(h)^T$ is also continuous and in particular the spectral radius $\rho(B(h))$ of $B(h)$ is continuous in $h$. In particular, for any $\epsilon>0$, there is some $\delta >0$ such that $|\rho(B(h))-\rho(B(0))|<\epsilon$. Finally, note that $\|A(h)\|= \sqrt{\rho(B(h))}$ to conclude the existence of $K$.