Statement in Riehl's book of the result "a diagram of functors has a limit if it has objectwise limit" $\def\A{\mathsf{A}}
\def\C{\mathsf{C}}
\def\ob{\operatorname{ob}}
\def\ev{\operatorname{ev}}
\def\J{\mathsf{J}}
$
I am having a little trouble understanding the last sentence of this result in p. 93 of Category Theory in Context, by E. Riehl.

Proposition 3.3.9. If $\A$ is small, then the forgetful functor $\C^\A\to\C^{\ob\A}$ strictly creates all limits and colimits that exist in $\C$. These limits are defined objectwise, meaning that for each $a\in\A$, the evaluation functor $\ev_a:\C^\A\to\C$ preserves all limits and colimits existing in $\C$.

(Italics are mine.)
I've read the proof of the result and completed the remaining details that the author leaves to the reader, but I'm still not sure what the phrase “the evaluation functor $\ev_a:\C^\A\to\C$ preserves all limits and colimits existing in $\C$” exactly means. Who is “all (co)limits existing in $\C$” in this phrase? It is false that $\ev_a$ preserves arbitrary limits (see here).
 A: $\def\A{\mathsf{A}}
\def\C{\mathsf{C}}
\def\ob{\operatorname{ob}}
\def\ev{\operatorname{ev}}
\def\J{\mathsf{J}}
$
I think the fine way to formulate what the last sentence is trying to say is:

The evaluation functor $\ev_a:\C^\A\to\C$ preserves all limits that were created pulling back along the forgetful functor $F:\C^\A\to\C^{\ob\A}$. More precisely, if $K:\J\to\C^\A$ is a diagram such that there is a limit cone over $FK$, then the unique lift along $\C^\A\to\C^{\ob\A}$ to a limit cone over $K$ is mapped through $\ev_a:\C^\A\to\C$ to a limit cone over $\ev_aK$.

The lift is unique since $F$ was shown to strictly create limits (Definition 3.3.7 in Riehl's book).
Proof. One could immediately arrive to the conclusion inspecting the limit construction in $\C^\A$ from the limit in $\C^{\ob\A}$. This explicit description of the limit is worked out in the proof that $F$ strictly creates limits.
Here is an alternative argument using diagramatic reasoning: There is a commutative diagram

where $\pi_a$ is the projection on the component $a\in\ob \A$. It can be shown that a diagram $\prod_{a\in\operatorname{ob}\mathsf{A}}K_a:\J\to\prod_{a\in\operatorname{ob}\mathsf{A}}\mathsf{C}$ has a limit if and only if all the diagrams $K_a:\J\to\C$, $a\in\ob\A$, have a limit and
$$
\lim_{\mathsf{J}}\prod_{a\in\operatorname{ob}\mathsf{A}}K_a
=\prod_{a\in\operatorname{ob}\mathsf{A}}
\lim_{\mathsf{J}}K_a.
$$
In particular, the projection $\pi_a$ preserves limits.
Therefore, if $K:\J\to\C$ is such that $FK$ has a limit cone, this will be mapped to another limit cone via the composite
\begin{equation}
\tag{1}\label{eq}
\C^{\ob\A}\xrightarrow{\cong}\prod_{\ob\A}\C\xrightarrow{\pi_a}\C
\end{equation}
of limit preserving functors. By commutativity of the diagram, mapping a limit cone along the composite \eqref{eq} is the same as first lifting it along $F$ and then mapping it through $\ev_a$. $\square$
