In example III.9.8.3, Hartshorne considers the projection $\varphi:\mathbb{P}^{n+1}\setminus P\to\mathbb{P}^n$ (projective space over an algebraic closed field $k$) where $P=(0,\ldots,0,1)$ ie $\varphi(x_0,\ldots,x_{n+1})=(x_0,\ldots,x_n)$ and for each $a\in k\setminus\{0\}$ the automorphism $\sigma_a$ of $\mathbb{P}^{n+1}$ is defined by $\sigma_a(x_0,\ldots,x_{n+1})=(x_0,\ldots,ax_{n+1})$.
Then he considers a closed subscheme $X_1$ of $\mathbb{P}^{n+1}$, not containing $P$ and sets for each $a\in k\setminus\{0\}$, $X_a=\sigma_a(X_1)$. He explains why we can consider the $X_a$ as a flat family over $\mathbb{A}^1\setminus\{0\}$.
We can then take the limit of this family in $0\in\mathbb{A}^1$, which is called $X_0$. Finally he says that set theortically one has $X_0=\varphi(X_1)$.
Question: why $X_0=\varphi(X_1)$ in terms of sets?
Ideas: consider for simplicity that we are in $\mathbb{A}^{n+1}$. If $X_1=V(I_1)=V(f_i)_i$ we have $X_a=V(f_i(x_1,\ldots,x_{n+1}/a))$ and the flat family is $X=V(f_i(x_1,\ldots,x_{n+1}/a))$ in $\mathbb{A}^{n+1}\times(\mathbb{A}^1\setminus\{0\})$ that is $X=V(f_i(x_1,\ldots,x_{n+1}b),ab-1)_i$ in $\mathbb{A}^{n+3}$. Then taking the schematic closure of $X$ in $\mathbb{A}^{n+1}\times\mathbb{A}^1$ is the same of eliminating $b$ ie $X=V(I)$ with $I=(f_i(x_1,\ldots,x_{n+1}b),ab-1)_i\cap k[x_1,\ldots,x_{n+1},a]$. Then $X_0$ is defined by $I_0=g_j(x_1,\ldots,x_{n+1},0)$ if $I=(g_j)$. On the other hand $\varphi(X_1)$ is defined by $J=I_1\cap k[x_1,\ldots,x_n]$ ie by eliminating $x_{n+1}$. So in the unlikely case of all this is correct, the question is why do we have: $$ ((f_i(x_1,\ldots,x_{n+1}/a))_i\cap k[x_1,\ldots,x_{n+1},a])_{|a=0}=(f_i(x_1,\ldots,x_{n+1}))_i\cap k[x_1,\ldots,x_n] $$