Basic example of a formal affine scheme, functorial point of view $\let\opn=\operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way. 
He defines (4.1) an formal scheme as a functor $X: \opn{CRings}\to \opn{Set}$ that is a small filtered colimit of affine schemes i.e., $X(R)=\lim\limits_{\rightarrow i}X_i(R)$. 
The first example (4.2) is given by the functor $\widehat{\mathbb {A}}^{1}   $ defined as $\widehat{\mathbb {A}}^{1}(R)\mathrel{:=}\opn{Nil}(R)$.
I don't understand why this functor is the colimit over $N$  of the functors $\opn{spec}(\mathbb{Z}[x]/x^{N+1})\mathrel{:=}\opn{Hom}_{\opn{CRing}}(\mathbb{Z}[x]/x^{N+1},\_)$.
I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!
 A: It might be illuminating to first work the example of (ordinary) affine space $\mathbb{A}^1_\mathbb{Z}$ over the integers.
As a functor, $\mathbb{A}^1_\mathbb{Z}$ is the forgetful functor $\mathit{Rings}^\mathrm{op}\rightarrow\mathit{Sets}$, sending a ring $R$ to its underlying set. It is representable by $\mathbb{Z}[t]$, as can be seen by the isomorphism $$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\cong R.$$
(As a homomorphism $f\colon\mathbb{Z}[t]\rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\rightarrow R$ by $f\mapsto f(t)$ for all $f\in\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)$.)
$\widehat{\mathbb{A}}^1_\mathbb{Z}$ is similar: there is an isomorphism
$$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t]/(t^n),R)\cong\mathrm{Nil}_n(R),$$
where $\mathrm{Nil}_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $\widehat{\mathbb{A}}^1_\mathbb{Z}$ sends $R$ to the colimit $\mathrm{colim}(\mathrm{Nil}_n(R))=\bigcup_{n\geq0}\mathrm{Nil}_n(R)=\mathrm{Nil}(R)$.
