Definition: A matrix lie group G is a group of matrices that is closed under nonsingular limits. That is, if A1, A2, A3,... is a convergent sequence of matrices in G, with limit A, and if detA is not 0, then A is in G.
Suppose we take any sequence in Aff(n) such that its limit is an element of $ GL_n(\mathbb{R})$. How do I show that (informally) the limit is in Aff(n)?
Aff(n) is the set of all matrices of the form $$ \begin{pmatrix} A & \ b \\ 0 & 1 \\ \end{pmatrix}$$ where A $\in GL_n(\mathbb{R})$, b in $\mathbb{R}^n$, and $0$ is 1 x n zero vector
Note that Aff(n) is a subset of n + 1 by n + 1 matrices
Your group is isomorphic to the subgroup of $GL(n+1, \mathbb R)$ of matrices of the form $$\begin{pmatrix}A&b\\0&1\end{pmatrix}$$ where $A$, $b$, $0$ and $1$ are blocks of sizes $n\times b$, $n\times 1$, $1\times n$ and $1\times 1$, respectively, by an obvious map.
Now this subgroup of $GL(n+1,\mathbb R)$ is the preimage of the vector $(0,\dots,0,1)\in\mathbb R^{n+1}$ under the function $\phi:GL(n+1,\mathbb R)\to\mathbb R^{n+1}$ such that if $B$ is an element of $GL(n+1,\mathbb R)$ then $\phi(B)$ is the last row of $B$.
Since $\phi$ is continuous and $\mathbb R^{n+1}$ Hausdorff, it is clear that your group is a closed subgroup of $GL(n+1,\mathbb R)$. Moreover, you can check easily that the differential of $\phi$ has the correct rank, so in fact it follows at once that the group is in fact a manifold. Etc.