This is essentially taken from the proof of Proposition 2.4.5 p. 36 in these notes by Pierre Schapira.

Let me use the abbreviation
$$
L:=\lim_{\underset{i}{\longleftarrow}}\quad.
$$

Let $C$ and $D$ be categories, let $b:I^{op}\to C$, $i\mapsto b_i$, be a projective system, let $Lb$ be its limit (we assume it exists), let $F:C\to D$ be a functor, let $G:D\to C$ be its left adjoint (we assume it exists), and let $y$ be an object of $D$.

We have the following commuting squares of morphisms and isomorphisms, where the vertical arrows are induced by the $i$ th canonical projections:
$$
\begin{matrix}
D(y,FLb)&\simeq&C(Gy,Lb)\\
\downarrow&&\downarrow\\
D(y,Fb_i)&\simeq&C(Gy,b_i),
\end{matrix}
$$

$$
\begin{matrix}
C(Gy,Lb)&\simeq&LC(Gy,b)\\
\downarrow&&\downarrow\\
C(Gy,b_i)&=&C(Gy,b_i),
\end{matrix}
$$

$$
\begin{matrix}
LC(Gy,b)&\simeq&LD(y,Fb)\\
\downarrow&&\downarrow\\
C(Gy,b_i)&\simeq&D(y,Fb_i),
\end{matrix}
$$

$$
\begin{matrix}
LD(y,Fb)&\simeq&D(y,LFb)\\
\downarrow&&\downarrow\\
D(y,Fb_i)&=&D(y,Fb_i).
\end{matrix}
$$
By splicing these squares, we get the following commuting square of morphisms and isomorphisms:
$$
\begin{matrix}
D(y,FLb)&\simeq&D(y,LFb)\\
\downarrow&&\downarrow\\
D(y,Fb_i)&=&D(y,Fb_i),
\end{matrix}
$$
which is what we wanted.

**EDIT A.** Let's go back to the first square:
$$
\begin{matrix}
D(y,FLb)&\simeq&C(Gy,Lb)\\
\downarrow&&\downarrow\\
D(y,Fb_i)&\simeq&C(Gy,b_i).
\end{matrix}
$$
The isomorphisms are given by the adjunction. If $p_i:Lb\to b_i$ denotes the $i$ th canonical projection, then the first downward arrow is $D(y,Fp_i)$, and the second is $C(Gy,p_i)$. To show that the square commutes, we only need to invoke the fact that the adjunction is functorial in the second variable.

Now to the second square:
$$
\begin{matrix}
C(Gy,Lb)&\simeq&LC(Gy,b)\\
\downarrow&&\downarrow\\
C(Gy,b_i)&=&C(Gy,b_i).
\end{matrix}
$$
By assumption, we have chosen a representing object $Lb$ and an isomorphism
$$
C(x,Lb)\simeq LC(x,b)
$$
functorial in $x\in\text{Ob}(C)$. We have a natural map from $LC(x,b)$ to $C(x,b_i)$ --- because $LC(x,b)$ is a projective limit of sets. Then we *define* the map from $C(x,Lb)$ to $C(x,b_i)$ as the one which makes the above square commutative. All this being functorial in $x$, the Yoneda Lemma yields the morphism $p_i:Lb\to b_i$ used above.

**EDIT B.** The third and fourth squares are handled similarly. So we end up with the square
$$
\begin{matrix}
D(y,FLb)&\simeq&D(y,LFb)\\
\downarrow&&\downarrow\\
D(y,Fb_i)&=&D(y,Fb_i),
\end{matrix}
$$
which commutes for all $i$. What we want to prove is the existence of an isomorphism $FLb\simeq LFb$ such that the square
$$
\begin{matrix}
FLb&\simeq&LFb\\
\downarrow&&\downarrow\\
Fb_i&=&Fb_i
\end{matrix}
$$
commutes for all $i$. But, in view of Yoneda, the above square commutes because the previous one does.

**EDIT C.** Alternative wording of the poof that
$$
\begin{matrix}
C(x,Lb)&\simeq&LC(x,b)\\
\downarrow&&\downarrow\\
C(x,b_i)&=&C(x,b_i)
\end{matrix}
$$
commutes:

A morphism $f\in C(x,Lb)$ is given by a family $f_\bullet=(f_j)_{j\in I}\in LC(x,b)$ satisfying the obvious compatibility conditions, and we have $f_j=p_j\circ f$ for all $j$. So, $f$ and $f_\bullet$ correspond under the isomorphism in the above square. Moreover, the first vertical arrow maps $f$ to $f_i$, and the second vertical arrow maps $f_\bullet$ to $f_i$.