Let $\mathsf{C}, \mathsf{D}$ be categories where all limits of shape $\mathsf{J}$ exist; let $G : \mathsf{C} \to \mathsf{D}$ be any functor. I am wondering why the canonical map $$\tau_F : [G \lim F \to \lim (G \circ F)] \in \mathsf{D}$$ is natural in $F : \mathsf{J} \to \mathsf{C}$. The map is defined by applying $G$ to the cone, $$[\lim F \to F] \overset{G}{\leadsto} [G \lim F \to G \circ F],$$ and then factoring according to the universal property of $\lim(G \circ F)$. This is OK with me. For naturality, I tried examining the square that should commute (morphism of diagrams $\eta : F_1 \to F_2$): $$\begin{array}{ccc} G \lim F_1 &\overset{\tau_{F_1}}{\to} & \lim (G \circ F_1) \\ \scriptstyle{G\lim \eta}\textstyle\downarrow\phantom{\scriptstyle{G \lim \eta}} & & \phantom{\scriptstyle{\lim G\eta}} \downarrow \scriptstyle{\lim G\eta} \\ G \lim F_2 & \underset{\tau_{F_2}}{\to} & \lim(G\circ F_2) \end{array}$$
But, it's not clear to me why it does. For instance, $\tau_F$ is defined abstractly, so I don't see how to just compute both directions and show that they are equal. On the other hand, $G \lim F_2$ doesn't have the universal property, so $(\lim G\eta) \circ \tau_{F_1}$ doesn't factor through $G \lim F_2$ in such a manner. Please let me know if I've misunderstood or overlooked something.
I believe this is an answer in line with Alessandro's hint. Fix $j \in \text{ob}(\mathsf{J})$ and append to the square the projections to $j$:
The two sectors commute by definition of $\tau$; the trapezoid commutes by definition of $\lim G\eta$; the outermost four arrows, by $G \lim \eta$. It follows that the paths $\pi_j \circ (\lim G \eta) \circ \tau_{F_1}$ and $\pi_j \circ \tau_{F_2} \circ (G \lim \eta)$ are equal. So by uniqueness in the universal property (Alessandro's hint), the original square commutes.