 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 limitFor $f:C\to C^{\prime }$, the arrows $D_j(f)p_{j,C}$ form a cone over a diagram at $C^{\prime }$, because the diagram arrows $D_j\to D_k$ are natural transformations. There is therefore a unique arrow $L(f):L(C)\to L(C^{\prime })$ withThe 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 $f:C\to C^{\prime }$, compare $L(f)q_C$ and $q_{C^{\prime }}X(f)$ after each $p_{j,C^{\prime }}$; 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 $[\mathcal{C},\mathcal{S}]$.

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.