[proofplan] We use only the universal property of the two limiting cones. Since $(L',(p'_j))$ is terminal among cones over $D$, the cone $(L,(p_j))$ admits a unique cone morphism $\varphi: L \to L'$. Reversing the roles of the two limits gives a unique cone morphism $\psi: L' \to L$. The composites $\psi \circ \varphi$ and $\varphi \circ \psi$ are forced to be identity morphisms by uniqueness in the corresponding terminal cone properties, so $\varphi$ is an isomorphism; uniqueness of $\varphi$ is the same universal property used to construct it. [/proofplan] [step:Use the terminal property of $L'$ to construct the comparison morphism] Because $(L',(p'_j)_{j \in \operatorname{Ob}(\mathcal{J})})$ is a limit of $D$, it is terminal in the category of cones over $D$. The family $(p_j)_{j \in \operatorname{Ob}(\mathcal{J})}$ makes $L$ into a cone over $D$. Therefore there exists a unique morphism \begin{align*} \varphi &: L \to L' \end{align*} in $\mathcal{C}$ such that, for every object $j \in \operatorname{Ob}(\mathcal{J})$, \begin{align*} p'_j \circ \varphi = p_j. \end{align*} [guided] The first limiting cone is \begin{align*} (L,(p_j)_{j \in \operatorname{Ob}(\mathcal{J})}), \end{align*} and the second limiting cone is \begin{align*} (L',(p'_j)_{j \in \operatorname{Ob}(\mathcal{J})}). \end{align*} Since $(L',(p'_j))$ is a limit, it is a terminal cone over the functor $D: \mathcal{J} \to \mathcal{C}$. Terminality means that every cone over $D$ admits a unique morphism of cones into $(L',(p'_j))$. We apply this terminal property to the cone $(L,(p_j))$. A morphism of cones from $(L,(p_j))$ to $(L',(p'_j))$ is exactly a morphism \begin{align*} \varphi &: L \to L' \end{align*} in $\mathcal{C}$ satisfying the compatibility equations \begin{align*} p'_j \circ \varphi = p_j \end{align*} for every object $j \in \operatorname{Ob}(\mathcal{J})$. Hence terminality of $(L',(p'_j))$ gives both existence and uniqueness of such a morphism $\varphi$. [/guided] [/step] [step:Use the terminal property of $L$ to construct the reverse comparison morphism] Similarly, since $(L,(p_j)_{j \in \operatorname{Ob}(\mathcal{J})})$ is a limit of $D$, it is terminal in the category of cones over $D$. Applying its terminal property to the cone $(L',(p'_j)_{j \in \operatorname{Ob}(\mathcal{J})})$, there exists a unique morphism \begin{align*} \psi &: L' \to L \end{align*} in $\mathcal{C}$ such that, for every object $j \in \operatorname{Ob}(\mathcal{J})$, \begin{align*} p_j \circ \psi = p'_j. \end{align*} [/step] [step:Show that $\psi \circ \varphi$ is the identity on $L$] Consider the composite morphism \begin{align*} \psi \circ \varphi &: L \to L. \end{align*} For every object $j \in \operatorname{Ob}(\mathcal{J})$, associativity of composition in $\mathcal{C}$ and the defining equations for $\varphi$ and $\psi$ give \begin{align*} p_j \circ (\psi \circ \varphi) &= (p_j \circ \psi) \circ \varphi \\ &= p'_j \circ \varphi \\ &= p_j. \end{align*} Thus $\psi \circ \varphi$ is a cone morphism from $(L,(p_j))$ to itself. The identity morphism \begin{align*} \operatorname{id}_L &: L \to L \end{align*} also satisfies \begin{align*} p_j \circ \operatorname{id}_L = p_j \end{align*} for every object $j \in \operatorname{Ob}(\mathcal{J})$. Since $(L,(p_j))$ is terminal in the category of cones over $D$, there is a unique cone morphism from $(L,(p_j))$ to itself. Therefore \begin{align*} \psi \circ \varphi = \operatorname{id}_L. \end{align*} [guided] We now prove that the two comparison maps are inverse in one direction. Define the composite morphism \begin{align*} \psi \circ \varphi &: L \to L. \end{align*} To show that this composite is the identity, we use the uniqueness part of the terminal property of the limiting cone $(L,(p_j))$. First we verify that $\psi \circ \varphi$ is a morphism of cones from $(L,(p_j))$ to itself. For each object $j \in \operatorname{Ob}(\mathcal{J})$, the defining property of $\psi$ gives $p_j \circ \psi = p'_j$, and the defining property of $\varphi$ gives $p'_j \circ \varphi = p_j$. Therefore, using associativity of composition in $\mathcal{C}$, \begin{align*} p_j \circ (\psi \circ \varphi) &= (p_j \circ \psi) \circ \varphi \\ &= p'_j \circ \varphi \\ &= p_j. \end{align*} So $\psi \circ \varphi$ preserves all cone structure maps. The identity morphism \begin{align*} \operatorname{id}_L &: L \to L \end{align*} also preserves all cone structure maps, since \begin{align*} p_j \circ \operatorname{id}_L = p_j \end{align*} for every object $j \in \operatorname{Ob}(\mathcal{J})$. Hence both $\psi \circ \varphi$ and $\operatorname{id}_L$ are cone morphisms from the cone $(L,(p_j))$ to itself. Because $(L,(p_j))$ is terminal in the category of cones over $D$, there is exactly one such cone morphism. Consequently, \begin{align*} \psi \circ \varphi = \operatorname{id}_L. \end{align*} [/guided] [/step] [step:Show that $\varphi \circ \psi$ is the identity on $L'$] Consider the composite morphism \begin{align*} \varphi \circ \psi &: L' \to L'. \end{align*} For every object $j \in \operatorname{Ob}(\mathcal{J})$, associativity of composition in $\mathcal{C}$ and the defining equations for $\varphi$ and $\psi$ give \begin{align*} p'_j \circ (\varphi \circ \psi) &= (p'_j \circ \varphi) \circ \psi \\ &= p_j \circ \psi \\ &= p'_j. \end{align*} Thus $\varphi \circ \psi$ is a cone morphism from $(L',(p'_j))$ to itself. The identity morphism \begin{align*} \operatorname{id}_{L'} &: L' \to L' \end{align*} also satisfies \begin{align*} p'_j \circ \operatorname{id}_{L'} = p'_j \end{align*} for every object $j \in \operatorname{Ob}(\mathcal{J})$. Since $(L',(p'_j))$ is terminal in the category of cones over $D$, there is a unique cone morphism from $(L',(p'_j))$ to itself. Therefore \begin{align*} \varphi \circ \psi = \operatorname{id}_{L'}. \end{align*} [/step] [step:Conclude that the comparison morphism is the unique required isomorphism] The equations \begin{align*} \psi \circ \varphi &= \operatorname{id}_L, \\ \varphi \circ \psi &= \operatorname{id}_{L'} \end{align*} show that $\varphi: L \to L'$ is an isomorphism with inverse $\psi: L' \to L$. By construction, $\varphi$ satisfies \begin{align*} p'_j \circ \varphi = p_j \end{align*} for every object $j \in \operatorname{Ob}(\mathcal{J})$. Its uniqueness among morphisms satisfying these equations was already obtained from the terminal property of the limiting cone $(L',(p'_j))$. Therefore there exists a unique isomorphism $\varphi: L \to L'$ with $p'_j \circ \varphi = p_j$ for every object $j \in \operatorname{Ob}(\mathcal{J})$. [/step]