Relation between limits and adjoints Let $\mathbf C$ be a complete category. For any diagram $J:\mathbf  I\to \mathbf C$, the functor $\operatorname{lim}(\mathbf C(-,J)):\mathbf C^{\operatorname{op}}\to\mathbf{Set}$ is representable, where $\mathbf C(-,J):\mathbf C\to [\mathbf I,\mathbf C]$ calculated on an object $C$ is the diagram $\mathbf C(C,J(-)):\mathbf I\to\mathbf{Set}$; call $\operatorname {lim}J$ the representing object, so that we have a natural isomorphism $\mathbf C(-,\operatorname {lim}J)\to\operatorname{lim}(\mathbf C(-,J))$. Together with the fact that the limit of a presheaf is computed objectwise, this means exactly that the Yoneda embedding preserves limits; plus, the fact that $\mathbf C(C,-)$ preserves limits, for any object $C$, becomes definitory.
$\operatorname{lim}(\mathbf C(C,J))\cong \operatorname{Nat}(\operatorname c(C),J)$,  naturally in $C$, where $\operatorname c(-):\mathbf C\to [\mathbf I,\mathbf {C}]$ is the functor sending $C$ to the constant functor on $C$; hence $\mathbf C(C,\operatorname{lim}(J))\cong \operatorname{Nat}(\operatorname c(C),J)$ naturally in $C$: does it already imply that $\operatorname{lim}$ can be extended (on arrows) to a functor $[\mathbf I,\mathbf {C}]\to\mathbf C$? I would say yes, because the universal property of the left adjoint $\operatorname c$, i.e. the naturality in $C$ of the latter isomorphism, should guarantee that. If I understood,  this could even be used as the definition of limit; sometimes it is referred to as a global  definition, I guess in the sense that it regards $\operatorname{lim}$ as a functor, on diagrams of shape $\mathbf  I$, instead of considering the diagram fixed.
Notice that in order for the first paragraph to make sense, we need to define preliminarly limits in $\mathbf {Set}$. In particular we could set that $\operatorname {lim}(D)$, for a diagram $\mathbf I\to \mathbf {Set}$, is equal to $\operatorname {Nat}(\operatorname c(\{*\},D)$; indeed, a posteriori, by the previous arguments $$\operatorname {Nat}(\operatorname c(\{*\},D)\cong \mathbf {Set}(\{*\},\operatorname{lim}(D))\cong \operatorname{lim}(D).$$
I tried to be as plain and clear as I could; I'm aware that this question may  be a triviality,  because I see this type of results often mentioned, but always taken for granted. The two\three textbooks that I looked at didn't make this point of  view explicit, although they always seem to touch it, so I'd like to know if this is actually how things work. Thank you in advance
 A: $\require{AMScd}$There are many ways in which you can prove that the assignment $D\mapsto \lim D$ is a functor $[I,C] \to C$; you seem to ask if there is a way to do it relying only on the definition of $\lim J$ as a representing object:

$\mathbf C(C,\lim D)\cong [{\bf I},{\bf C}](\operatorname c(C),D)$ naturally in $C$

Your question is:

does it already imply that $\operatorname{lim}$ can be extended (on arrows) to a functor $[\mathbf I,\mathbf {C}]\to\mathbf C$?

Given a natural transformation $\alpha : D \Rightarrow D'$, you have three arrows like these:
$$\begin{CD}
{\bf C}(C, \lim D) @>\cong>t_{C,D}> [{\bf I},{\bf C}](\operatorname c(C),D) \\ 
@. @VV[{\bf I},{\bf C}](\operatorname c(C),\alpha)V \\
{\bf C}(C, \lim D') @>t_{C,D'}>\cong> [{\bf I},{\bf C}](\operatorname c(C),D') 
\end{CD}$$
which gives a composite map
$$ \begin{CD} {\bf C}(C, \lim D) @>>> [{\bf I},{\bf C}](\operatorname c(C),D) @>>> [{\bf I},{\bf C}](\operatorname c(C),D')  @>>> {\bf C}(C, \lim D')\end{CD} $$
natural in $C$. But this means that there is a natural transformation between the representable presheaves ${\bf C}(\_, \lim D)$ and ${\bf C}(\_, \lim D')$, which must come from a unique morphisms in $\bf C$ between $\lim D$ and $\lim D'$, because the Yoneda embedding is fully faithful.
The "unique" part gives you that this action on morphisms is functorial, and since this is very standard and instructive, I'll leave it to you.
Certainly, the limit of $D$ can also be defined in other ways; I agree to call "global" the definition that says that $D\mapsto \lim D$ is right adjoint to the constant functor because it is strictly stronger than asking just that a single limit for a single diagram $D$ exists.
