Let $k$ be a field. Then consider the rings $k[T] / (T^n)$ with $n \in \mathbb{N}$. The inverse limit of these is given by $k[[T]]$. Passing on to the category of schemes, one concludes that the direct limit of $\text{Spec}(k[T] / (T^n)) $ is given by $\text{Spec}(k[[T]]) $.
Now to the question. $\text{Spec}(k[T] / (T^n)) $ can be viewed as one point on the affine line with an infinitisimal neighborhood that remembers some derivatives of functions, right?
As the direct limit, $\text{Spec}(k[[T]]) $ is one point on the affine line, that remembers all derivatives of functions. But this scheme has two points, one being a specialisation of the other.
Are there any intuitive geometric explanations for this phenomena?



Question concerning $\text{Spec}(k[[T]]) $
I'm not sure if you''re looking for something deeper than this (and therefore would have preferred to make this a comment if I could have squeezed it in), but:


$Spec(A)$ has to be just rich enough so that every map from $A$ to a field shows up as a function on $Spec(A)$.


Now the issue is that, for a field $F$, the functor $Hom\_{\bf Rings}(-,F)$ does not preserve inverse limits. So if $A$ is the inverse limit of rings $A\_n$, you can't in general understand maps $A\rightarrow F$ as limits of maps $A\_n\rightarrow F$, which means you shouldn't expect to understand the points of $Spec(A)$ as limits of points of $Spec(A\_n)$.


In particular, put $A\_n=Spec\Big(k[[t]]/(t^n)\Big)$ and $A=k[[t]]=\lim\_\leftarrow A\_n$. Put $F=A[t^{-1}]$. Then there is a ring map $A\rightarrow F$ (namely the obvious inclusion) that does not arise from maps $A\_n\rightarrow F.$ This necessitates a point in $Spec(A)$ that does not arise from points in the various $Spec(A\_n)$.


