[proofplan] We prove the universal property of the displayed cone over $G\circ D$. Given any cone from an object $X\in\mathcal C$ to $G\circ D$, we transpose its legs across the adjunction to obtain a cone from $F(X)$ to $D$. The limiting property of $L$ gives a unique map $F(X)\to L$, and transposing this map back gives the required unique map $X\to G(L)$. Naturality and bijectivity of the adjunction bijections verify both the cone equations and uniqueness. [/proofplan] [step:Fix the adjunction bijections and the induced cone over $G\circ D$] Since $F\dashv G$, for every object $X\in\mathcal C$ and every object $Y\in\mathcal D$ there is a natural bijection \begin{align*} \Phi_{X,Y}:\operatorname{Hom}_{\mathcal C}(X,G(Y))\to \operatorname{Hom}_{\mathcal D}(F(X),Y). \end{align*} For each object $j\in\operatorname{Ob}(J)$, the morphism $\lambda_j:L\to D(j)$ in $\mathcal D$ gives a morphism \begin{align*} G(\lambda_j):G(L)\to G(D(j)) \end{align*} in $\mathcal C$. If $a:j\to k$ is a morphism in $J$, then the cone equation for $(L,(\lambda_j))$ says \begin{align*} D(a)\circ \lambda_j=\lambda_k. \end{align*} Applying the functor $G$ gives \begin{align*} G(D(a))\circ G(\lambda_j)=G(D(a)\circ \lambda_j)=G(\lambda_k), \end{align*} so $\left(G(L),(G(\lambda_j))\right)$ is a cone over $G\circ D$. [/step] [step:Transpose an arbitrary cone to a cone in $\mathcal D$] Let $X\in\operatorname{Ob}(\mathcal C)$, and let \begin{align*} \alpha_j:X\to G(D(j)) \end{align*} for $j\in\operatorname{Ob}(J)$ be a cone from $X$ to $G\circ D$. Thus, for every morphism $a:j\to k$ in $J$, \begin{align*} G(D(a))\circ \alpha_j=\alpha_k. \end{align*} For each $j\in\operatorname{Ob}(J)$, define the adjoint transpose \begin{align*} \beta_j:F(X)\to D(j) \end{align*} by \begin{align*} \beta_j:=\Phi_{X,D(j)}(\alpha_j). \end{align*} We claim that $(\beta_j)_{j\in\operatorname{Ob}(J)}$ is a cone from $F(X)$ to $D$. Let $a:j\to k$ be a morphism in $J$. Naturality of $\Phi$ in the second variable, applied to the morphism $D(a):D(j)\to D(k)$, gives \begin{align*} \Phi_{X,D(k)}(G(D(a))\circ \alpha_j)=D(a)\circ \Phi_{X,D(j)}(\alpha_j). \end{align*} Using the cone equation for $\alpha$ and the definition of $\beta_j$, this becomes \begin{align*} \Phi_{X,D(k)}(\alpha_k)=D(a)\circ \beta_j. \end{align*} Since $\beta_k=\Phi_{X,D(k)}(\alpha_k)$, we obtain \begin{align*} D(a)\circ \beta_j=\beta_k. \end{align*} Thus $(\beta_j)$ is a cone from $F(X)$ to $D$. [/step] [step:Use the limiting property of $L$ and transpose back] Since $(L,(\lambda_j))$ is a limit cone for $D$, the cone $(\beta_j)_{j\in\operatorname{Ob}(J)}$ from $F(X)$ to $D$ factors through $L$ by a unique morphism \begin{align*} v:F(X)\to L \end{align*} such that, for every $j\in\operatorname{Ob}(J)$, \begin{align*} \lambda_j\circ v=\beta_j. \end{align*} Define \begin{align*} u:X\to G(L) \end{align*} to be the inverse adjoint transpose of $v$: \begin{align*} u:=\Phi_{X,L}^{-1}(v). \end{align*} We verify that $u$ induces the original cone $(\alpha_j)$. For each $j\in\operatorname{Ob}(J)$, naturality of $\Phi$ in the second variable, applied to $\lambda_j:L\to D(j)$, gives \begin{align*} \Phi_{X,D(j)}(G(\lambda_j)\circ u)=\lambda_j\circ \Phi_{X,L}(u). \end{align*} Because $\Phi_{X,L}(u)=v$, the right-hand side is \begin{align*} \lambda_j\circ v=\beta_j=\Phi_{X,D(j)}(\alpha_j). \end{align*} Hence \begin{align*} \Phi_{X,D(j)}(G(\lambda_j)\circ u)=\Phi_{X,D(j)}(\alpha_j). \end{align*} Since $\Phi_{X,D(j)}$ is a bijection, it follows that \begin{align*} G(\lambda_j)\circ u=\alpha_j. \end{align*} Thus $u:X\to G(L)$ is a mediating morphism from $X$ to the cone $\left(G(L),(G(\lambda_j))\right)$. [/step] [step:Prove uniqueness of the mediating morphism] Let \begin{align*} u':X\to G(L) \end{align*} be any morphism such that, for every $j\in\operatorname{Ob}(J)$, \begin{align*} G(\lambda_j)\circ u'=\alpha_j. \end{align*} Define its adjoint transpose \begin{align*} v':F(X)\to L \end{align*} by \begin{align*} v':=\Phi_{X,L}(u'). \end{align*} For each $j\in\operatorname{Ob}(J)$, naturality of $\Phi$ in the second variable gives \begin{align*} \lambda_j\circ v' &=\lambda_j\circ \Phi_{X,L}(u')\\ &=\Phi_{X,D(j)}(G(\lambda_j)\circ u')\\ &=\Phi_{X,D(j)}(\alpha_j)\\ &=\beta_j. \end{align*} Thus $v':F(X)\to L$ is another morphism inducing the cone $(\beta_j)$ through the limit cone $(L,(\lambda_j))$. By uniqueness in the universal property of $L$, we have \begin{align*} v'=v. \end{align*} Applying the inverse of the bijection $\Phi_{X,L}$ gives \begin{align*} u'=\Phi_{X,L}^{-1}(v')=\Phi_{X,L}^{-1}(v)=u. \end{align*} Therefore the mediating morphism $u:X\to G(L)$ is unique. Since this holds for every object $X\in\mathcal C$ and every cone from $X$ to $G\circ D$, the cone \begin{align*} \left(G(L),\left(G(\lambda_j):G(L)\to G(D(j))\right)_{j\in\operatorname{Ob}(J)}\right) \end{align*} satisfies the universal property of a limit cone for $G\circ D$ in $\mathcal C$. [/step]