node {$\tilde p_D$} (m-2-2);
\draw[->] (m-2-1) -- node [swap]{$\tilde p_C$} (m-2-2);
\end{tikzpicture}$$ where $\lambda_p=\lambda_C\times_K\lambda_D:{\mathcal}{M}_p\to {\mathcal}{C}_p\times_K{\mathcal}{D}_p$ is a right fibration and $\tilde p_C$ and $\tilde p_D$ are Cartesian fibrations. The limit of $p$ is a right fibration (cf. [@Lurie:ZsNVU6Ru Remark 4.2.9.]) $$\label{eq:LimPair}{\mathcal}{M}:=\lim p_M\to \lim p_C\times \lim p_D=:{\mathcal}{C}\times{\mathcal}{D},$$ where $p_M:K\to {{\mathscr{C}\mathrm{at}}}_\infty$ is the functor classified by $(\tilde p_C\times_K\tilde p_D)\circ\lambda_p$, and $p_C$ and $p_D$ are classified by $\tilde p_C$ and $\tilde p_D$ respectively. We need to show that is right representable and that the canonical morphisms to are right representable.

Now a vertex $d\in \lim p_D={\mathcal}{D}$ can be identified with a Cartesian section $\tilde d:K\to {\mathcal}{D}_p$ of $\tilde p_D$ (cf. [@Lurie:2009un Corollary 3.3.3.2.]). Let ${\mathcal}{M}_{p,d}\xrightarrow{\lambda_d} K$ be the Cartesian fibration which fits into the pullback square $$\begin{tikzpicture}
    {\matrix (m) [matrix of math nodes, row sep=3em,
  column sep=3em, text height=1.5ex, text depth=0.25ex]}{{\mathcal}{M}_{p,d}& {\mathcal}{M}_p\\K& {\mathcal}{D}_p\\};
    \draw[->] (m-1-1) -- (m-1-2);
    \draw[->] (m-1-1) -- (m-2-1);
    \draw[->] (m-1-2) -- (m-2-2);
    \draw[->] (m-2-1) -- node {$\tilde d$} (m-2-2);
\end{tikzpicture}$$ and let $q:K^{op}\to {{\mathscr{C}\mathrm{at}}}_\infty$ be the corresponding functor. Then $\lim q\cong {\mathcal}{M}_d:={\mathcal}{M}\times_{{\mathcal}{D}}\{d\}$ (since taking pullbacks commutes with taking limits). To show that is right representable, we need to show that ${\mathcal}{M}_d$ has a final object. However, since $p$ takes values 