Limits in the category of Posets Let $\textbf{n}$ be the poset of natural numbers $0\leq1\leq2\leq\ldots\leq n$ (for $n\in\mathbb{N}$). 
Consider the sequence $$\textbf{0}\rightarrow\textbf{1}\rightarrow\textbf{2}\rightarrow\cdots$$ where each arrow is an inclusion. 
I want to compute the limit poset of this sequence in the category $\mathbf{Posets}$ of posets and monotone functions. 

What I have done so far: Intuitively the limit should be $(\mathbb{N},\leq)$. But let us go with the definition. Define the index category $\Bbb{J}$ by the set of natural numbers with arrow $\alpha_i:i\rightarrow i+1$ and define the diagram $D:\Bbb{J}\rightarrow\textbf{Posets}$ by $D(j)=\textbf{j}$ and $D_{\alpha_i}:D(i)\rightarrow D(i+1)$ by the given inclusions.
But now my problem: I don't know what the cones are and thus I don't know how to find a terminal object in $\textbf{Cone(D)}$ to find the limit. Can someone help me with this? :)
Thanks.
 A: Edit: I assumed without second thought that $\mathbf 0$ signified the empty poset. My answer should be read in this context.

$(\Bbb N,\le)$ is the colimit of that sequence. The limit (and the only cone) is $\mathbf 0$.

First, let us focus on the last statement. A cone is, by definition, a natural transformation $\mu:c_P \to D$, where $c_P$ is the constant functor at the object $P$ of $\mathsf{Posets}$ (think of $P$ as the apex of the cone, with the components $\mu_i$ as the edges connecting the apex to the base $D$).
That is, a collection of morphisms $\mu_i: P \to D(i) = \mathbf i$ such that $\mu_j = (i \to j)\circ\mu_i$. Since $\mathbf 0$ is the initial object of (the subcategory defined by) $D$, this condition implies that $\mu$ is completely determined by $\mu_0: P \to \mathbf 0$. But $\mathbf 0$ is strictly initial in the category $\mathsf{Posets}$, that is, it must be that $P \cong \mathbf 0$ in $\mathsf{Posets}$.
Hence, $\mathsf{Cone}(D) \cong \mathbf 1$ as categories, implying that the cone determined by ${\rm id}:\mathbf 0 \to \mathbf 0$ is the limit of $D$.

Dually, a cocone is a natural transformation $\nu: D \to C_Q$, with $c_Q$ the constant functor at the poset $Q$. That is, a collection of morphisms $\nu_i: \mathbf i = D(i) \to Q$ such that $\nu_i = \nu_j\circ(i \to j)$.
If you write it down explicitly, you'll see that the $\nu_i$ amount to an inductive definition of $\bar\nu: \Bbb N \to Q$. This you can apply to demonstrate that $(\Bbb N,\le)$ is indeed the colimit: the initial object of $\mathsf{Cocone}(D)$.
A: As I assume this exercise is taken from Awodey's book, where the notation is $[n]=\{0\leq 1\leq ... \leq n\}$ and the sequence is $[0]\to[1]\to...$, I will write it this way.
Lord_Farin's answer seems correct regarding the colimit part, as it should be $\mathbb N$. The other part however seems less right since the whole argument rests on the first arrow of the cone going to the empty set, which isn't the same as $[0]$.
With this in mind the limit should reasonably be the singleton poset itself. The empty set can't be a limit, since given another cone $(C,\lambda_n)$, with $C$ non-empty, there is no arrow to the empty set!
However, there is a unique arrow $C\to[0]$, and since the morphisms $\lambda_n$ already commute with inclusions in the sequence everything also commutes with them through the canonical isomorphism $[0]\to[0]$.
