Here is the question I want a concrete example of it so that I can feel that the given set is really a limit:
Show that the set $(5.16)$ in Example 5.1.22 really is a limit of $D.$
Here is Example 5.1.22:
Could someone help me please by giving me a concrete example?
You can take $\mathbf{I}$ to be "$\mathbf{\cdot} ~~~ \mathbf{\cdot}$", and define the diagram to be $D: \mathbf{I} \to \mathbf{Set}$ be $\{a, b\}$, $\{c, d\}$. (Just two set/object without map/morphism). Then according to the construction in (5.16), it's:
$$ \{(x, y)| x \in \{a, b\}, y \in \{c, d\}\text{ such that no more condition are need since no morphisms in } \mathbf{I}\} $$
Try to change $\mathbf{I}$ to be $\cdot \rightarrow \cdot$ may help too.
Here I skipped the argument for 5.16 and the reality check for it being a limit.
$\newcommand{\set}{\mathsf{Set}}$It's unclear what you want. I provide here an example of a non-trivial limit and a strong hint as to how to prove it is a limit. As there is only one arrow between any two objects in the diagram, I think this is simple enough to be understood as a first example whilst still being non-trivial.
The $p$-adic integers are a great example. I enjoyed reading this classic MSE post on the matter. I know little number theory, so I can't tell you why the (analytic?) number theorists get excited about $p$-adic number systems but they're a solid introductory example of a non-trivial limit system.
In the category $\set$ (you could consider the category of groups or rings too) define the object $\Bbb Z_n$, for $n\in\Bbb N$, as the ring of integers modulo $n$. Pick a prime $p$ and "construct" the following diagram:
$$\large\cdots\twoheadrightarrow\Bbb Z_{p^{k+1}}\overset{\pi_k}{\twoheadrightarrow}\Bbb Z_{p^k}\overset{\pi_{k-1}}{\twoheadrightarrow}\Bbb Z_{p^{k-1}}{\twoheadrightarrow}\cdots\twoheadrightarrow\Bbb Z_{p^2}\overset{\pi_1}{\twoheadrightarrow}\Bbb Z_p$$
Where the $\pi_k:\Bbb Z_{p^{k+1}}\twoheadrightarrow\Bbb Z_{p^k}$ are the surjective projections that assign to every $n$ modulo $p^{k+1}$ the element $n\mod p^k$. For instance, with $p=5$ and $k=2$, the integer $78$ modulo $5^3=125$ is projected to $3\equiv78$ modulo $5^2$. I've avoided equivalence class notation because I think it would be confusing here.
Why does it have to be prime? It doesn't, for the purpose of building a limit system, but the use of a prime allows the $p$-adic integers to enjoy special properties - see the use of Hensel's Lemma in the linked post.
The limit object is the $p$-adic integers. They're defined to have the elements: $$\{(n_1,n_2,\cdots):n_k\in\Bbb Z,\,n_{k+1}\equiv n_k\bmod p^k\,\forall k\in\Bbb N\}$$Notice that this is exactly Leinster's limit construction (I omit the condition $n_{k+m}\equiv n_k\mod p^k$ for all $k,m$ - that is, $(Du)x_i=x_j$ for all $u:i\to j$ - since this reduces to the single-step condition $n_{k+1}\equiv n_k$).
Why is it a limit? If $R\in\set$ has some family of functions $\varphi_k:R\to\Bbb Z_{p^k}$ for which $\pi_k\circ\varphi_{k+1}=\varphi_k$ for all $k$, then ... you can continue the proof. My hint: we know $\varphi_{k+1}(r)=\varphi_k(r)\bmod p^k$ for all $k$ and $r\in R$. How can we build a $p$-adic integer from this (for every $r$)?
