= Solution
Let $D:J\to[\mathcal C,\mathcal S]$ be a small <diagram in a category>. Since $\mathcal S$ is a <complete category>, choose at each $C\in\mathcal C$ a <categorical limit>
$$
L(C)=\lim_{j\in J}D_j(C),\qquad p_{j,C}:L(C)\to D_j(C).
$$
For $f:C\to C'$, the arrows $D_j(f)p_{j,C}$ form a <cone over a diagram> at $C'$, because the diagram arrows $D_j\to D_k$ are <natural transformations>. There is therefore a unique arrow $L(f):L(C)\to L(C')$ with
$$
p_{j,C'}L(f)=D_j(f)p_{j,C}.
$$
The <universal property> immediately gives $L(1_C)=1_{L(C)}$ and $L(gf)=L(g)L(f)$: both sides have the same composites with every limit projection. Thus $L$ is a <functor>, and the displayed equations make the $p_j$ <natural transformations>.
Given any <categorical cone> $q_j:X\to D_j$ in the <functor category>, its components induce unique $q_C:X(C)\to L(C)$. For $f:C\to C'$, compare $L(f)q_C$ and $q_{C'}X(f)$ after each $p_{j,C'}$; <naturality> of $q_j$ makes their composites equal. Hence $q$ is a <natural transformation>, and uniqueness is componentwise. This proves that $L$ is a <categorical limit> in $[\mathcal C,\mathcal S]$.
\b[The functor category is complete, and every evaluation functor preserves limits], since evaluating the constructed limit at $C$ gives exactly the chosen limit in $\mathcal S$. This is the <pointwise limits in a functor category> construction; it also covers the empty diagram and its <terminal object>.
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