[proofplan]
We prove the universal property of the cone $G(\lambda_j):G(L)\to G(D(j))$. Given any object $c\in\mathcal C$ and any cone from $c$ to $G\circ D$, we transpose each leg across the adjunction to obtain a cone from $F(c)$ to $D$. Since $L$ is the limit of $D$, that cone factors uniquely through $L$; transposing the resulting morphism $F(c)\to L$ back across the adjunction gives the unique morphism $c\to G(L)$ required for the limit property.
[/proofplan]
[step:Fix the adjunction bijections and the limiting cone]
For objects $c\in\mathcal C$ and $d\in\mathcal D$, let
\begin{align*}
\Phi_{c,d}:\operatorname{Hom}_{\mathcal D}(F(c),d) \to \operatorname{Hom}_{\mathcal C}(c,G(d))
\end{align*}
denote the adjunction bijection, natural in both variables. Let
\begin{align*}
\lambda_j:L\to D(j)
\end{align*}
for $j\in\operatorname{Ob}(J)$ be the given limiting cone. Since $G$ is a functor, the morphisms
\begin{align*}
G(\lambda_j):G(L)\to G(D(j))
\end{align*}
form a cone over $G\circ D$: for every morphism $u:j\to k$ in $J$,
\begin{align*}
G(D(u))\circ G(\lambda_j)
=
G(D(u)\circ \lambda_j)
=
G(\lambda_k),
\end{align*}
because $\lambda_j$ is a cone over $D$.
[/step]
[step:Transpose an arbitrary cone over $G\circ D$ to a cone over $D$]
Fix an object $c\in\mathcal C$. Let
\begin{align*}
\alpha_j:c\to G(D(j))
\end{align*}
for $j\in\operatorname{Ob}(J)$ be a cone from $c$ to $G\circ D$, meaning that for every morphism $u:j\to k$ in $J$,
\begin{align*}
G(D(u))\circ \alpha_j=\alpha_k.
\end{align*}
For each $j\in\operatorname{Ob}(J)$, define
\begin{align*}
\beta_j:F(c)\to D(j)
\end{align*}
to be the unique morphism satisfying
\begin{align*}
\Phi_{c,D(j)}(\beta_j)=\alpha_j.
\end{align*}
We claim that the family $(\beta_j)_{j\in\operatorname{Ob}(J)}$ is a cone from $F(c)$ to $D$. Let $u:j\to k$ be a morphism in $J$. By naturality of $\Phi_{c,-}$ in the $\mathcal D$-variable,
\begin{align*}
\Phi_{c,D(k)}(D(u)\circ \beta_j)
=
G(D(u))\circ \Phi_{c,D(j)}(\beta_j)
=
G(D(u))\circ \alpha_j
=
\alpha_k
=
\Phi_{c,D(k)}(\beta_k).
\end{align*}
Since $\Phi_{c,D(k)}$ is injective, it follows that
\begin{align*}
D(u)\circ \beta_j=\beta_k.
\end{align*}
Thus $(\beta_j)$ is a cone from $F(c)$ to $D$.
[/step]
[step:Use the limiting property of $L$ and transpose back]
Since $(\lambda_j:L\to D(j))$ is a limiting cone for $D$, there exists a unique morphism
\begin{align*}
h:F(c)\to L
\end{align*}
such that for every $j\in\operatorname{Ob}(J)$,
\begin{align*}
\lambda_j\circ h=\beta_j.
\end{align*}
Define
\begin{align*}
f:c\to G(L)
\end{align*}
by
\begin{align*}
f:=\Phi_{c,L}(h).
\end{align*}
For each $j\in\operatorname{Ob}(J)$, naturality of $\Phi_{c,-}$ gives
\begin{align*}
G(\lambda_j)\circ f
=
G(\lambda_j)\circ \Phi_{c,L}(h)
=
\Phi_{c,D(j)}(\lambda_j\circ h)
=
\Phi_{c,D(j)}(\beta_j)
=
\alpha_j.
\end{align*}
Therefore $f$ factors the cone $(\alpha_j)$ through the cone $(G(\lambda_j))$.
[/step]
[step:Prove the factorization through $G(L)$ is unique]
Let
\begin{align*}
f':c\to G(L)
\end{align*}
be any morphism such that for every $j\in\operatorname{Ob}(J)$,
\begin{align*}
G(\lambda_j)\circ f'=\alpha_j.
\end{align*}
Define
\begin{align*}
h':F(c)\to L
\end{align*}
to be the unique morphism satisfying
\begin{align*}
\Phi_{c,L}(h')=f'.
\end{align*}
For each $j\in\operatorname{Ob}(J)$, naturality of $\Phi_{c,-}$ gives
\begin{align*}
\Phi_{c,D(j)}(\lambda_j\circ h')
=
G(\lambda_j)\circ \Phi_{c,L}(h')
=
G(\lambda_j)\circ f'
=
\alpha_j
=
\Phi_{c,D(j)}(\beta_j).
\end{align*}
Since $\Phi_{c,D(j)}$ is injective,
\begin{align*}
\lambda_j\circ h'=\beta_j
\end{align*}
for every $j\in\operatorname{Ob}(J)$. By the uniqueness part of the limiting property of $L$, we have $h'=h$. Applying $\Phi_{c,L}$ yields
\begin{align*}
f'=\Phi_{c,L}(h')=\Phi_{c,L}(h)=f.
\end{align*}
Thus for every object $c\in\mathcal C$, every cone from $c$ to $G\circ D$ factors uniquely through the cone $(G(\lambda_j))$. Hence $(G(\lambda_j):G(L)\to G(D(j)))$ is a limiting cone for $G\circ D$ in $\mathcal C$.
[/step]
