I realize this is a duplicate of this question, but I do not understand the answer. I am talking about the contravariant Yoneda embedding and this is how far I got:
Let $D: I \to \mathcal{C}$ be a diagram and $(C, \sigma)$ be a limiting cone of $D$. For any $X \in \mathcal{C}$ we can define a functor $\hom(X, D-): I \to Set$ that sends $i$ to $\hom(X, Di)$ and $f: i \to j$ to $Df \circ -$. It is easy to see that this together with the natural transformation $(\sigma_i \circ -)_{i \in I}$ forms a cone by writing out the diagram and using $(C, \sigma)$ is a cone. I fail to see how this cone is limiting. For any set $K$ and natural transformation $\mu: \Delta_K \to \hom(X, D-)$ we would need to find a functions $\nu_i: K \to \hom(X, C)$ such that $(\sigma \circ - ) \circ \nu_i = \mu_i$ and hope that these bundle nicely together into a natural transformation $\nu$, but I do not even see how we can define a function from $K$ to $\hom(X, C)$.
Note on notation in case it is not standard: I use the notation $\sigma \circ -$ for the function that sends some function $\alpha$ to $\sigma \circ \alpha$.
Edit building on Arnaud's answer
For $\nu$ constructed from $X$ we write $\nu_X$.
Let $F$ be a functor and $(F, \kappa)$ be a cone for $yD$. From the construction of $\nu$ it follows that if $(\nu_X)_{X \in \mathcal{C}}$ is a natural transformation from $F$ to $\hom(-, C)$ we are done. I don't know how to do this. Let $g: B \to A$ be an arrow, then we have a diagram
where the large square commutes because $\kappa_J$ is a natural transformation, the right square commutes by just checking and we need to prove that the left square commutes. I cannot make it work. It would follow from $\sigma_J$ being mono, but that is not necessarily true. Maybe using that this is true for all $J \in I$ we can do something similar?
Suppose we are given a natural transformation $\mu: \Delta_K \to \hom(X, D-)$, i.e. a family of maps $\mu_i:K\to \hom(X,Di)$ commuting with all the maps $Df\circ - $. If you fix an element $k\in K$, then for every object $i$ of $I$ you have an arrow $\mu_i(k):X\to Di$ in $\mathcal{C}$, and $$Df\circ \mu_i(k)=(Df\circ -)(\mu_i(k))=((Df\circ -)\circ \mu_i)(k)=\mu_j(k)$$ for all $f:i\to j$ in $I$. In other words, the $\mu_i(k)$ forms a cone over $D$ in $\mathcal{C}$, and thus by the universal property you have an arrow $\nu(k):X\to C$ such that $\sigma_i\circ \nu (k)=\mu_i(k)$. This describes a function $\nu:K\to \hom(X,C)$, and by construction it has the property that $(\sigma_i\circ -)\circ \nu=\mu_i$ for all $i$. You can check that this $\nu$ is unique, and this shows that $\hom(X,C)$ is the limit of $\hom(X,D-)$. Since it is true for all $X$, this is sufficient to prove that $\hom(-,C)$ is the limit of $\hom(-,D-)$ in $[\mathcal{C}^{op},\mathbf{Set}]$.