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 $f_i$.