Proof.
We will prove the first assertion; the second follows by a similar argument. Projection onto the first factor determines a right fibration $\theta : \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{ \operatorname{Fun}(K,\operatorname{\mathcal{C}}) } \{ u\} \rightarrow \operatorname{\mathcal{C}}$. For each object $X \in \operatorname{\mathcal{C}}$, we can identify $\theta ^{-1}(X)$ with the morphism space $\operatorname{Hom}_{\operatorname{Fun}(K, \operatorname{\mathcal{C}})}( \underline{X}, u )$. Let
\[ \rho _{X}: \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{C}})}( \underline{Y}, u) \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X,Y) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(K,\operatorname{\mathcal{C}}) }( \underline{X}, u) \]
be the parametrized contravariant transport map of Variant 5.2.8.6. Using Remark 5.2.8.5 and Proposition 5.2.8.7, we see that $\rho _{X}$ factors as a composition
\begin{eqnarray*} \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{C}})}( \underline{Y}, u) \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X,Y) & \rightarrow & \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{C}})}( \underline{Y}, u) \times \operatorname{Hom}_{\operatorname{Fun}(K,\operatorname{\mathcal{C}})}( \underline{X}, \underline{Y} )\\ & \xrightarrow {\circ } & \operatorname{Hom}_{ \operatorname{Fun}(K,\operatorname{\mathcal{C}}) }( \underline{X}, u),\end{eqnarray*}
given on objects by the construction $( \alpha , f) \mapsto \alpha \circ \underline{f}$. It follows that a natural transformation $\alpha : \underline{Y} \rightarrow u$ exhibits $Y$ as a limit of $u$ if and only if, for every object $X \in \operatorname{\mathcal{C}}$, the restriction $\rho _{X}|_{ \{ \alpha \} \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) }$ is a homotopy equivalence of Kan complexes. By virtue of Proposition 5.6.6.21, this is equivalent to the requirement that $\alpha $ is final when regarded as an object of the $\infty $-category $\operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{ \operatorname{Fun}(K,\operatorname{\mathcal{C}}) } \{ u\} $.
$\square$