Via Theorem 3.90:

\$$V=\\{1\\}\$$. \$$\mathrm{lim}\_\textbf{1}D=\\{(d_1)|d\_1\in D(1)\,\mathrm{and}\,D(\mathrm{id}\_1)(d_1)=d_1\\}\$$ together with a projection map \$$p_1((d_1))=d_1\$$. However, since \$$D\$$ is a functor, the constraint is always satisfied, and the projection map is the identity. So, the limit of \$$D:\textbf{1}\to\textbf{Set}\$$ is the set that it picks out in **Set**.

Via Definition 3.87:

Let \$$S\in\textbf{Set}\$$ be the set that \$$D\$$ picks out. A cone over \$$D\$$ consists of a set \$$C\$$ and a function \$$f:C\to S\$$ (the cone property is trivial here, just like the constraint in theorem 3.90). Morphisms between these cones are functions \$$a:C\to C'\$$ such that for \$$g:C'\to S\$$, \$$g\circ a=f\$$. A terminal object in this category **Cone**(\$$D\$$) is a set \$$T\$$ and a function \$$h:T\to S\$$ such that for every set \$$C\$$ and function \$$f:C\to S\$$ there is a unique \$$b:C\to T\$$ with \$$h\circ b=f\$$. For each element \$$s\in S\$$ there is a cone \$$\\{ * \\},f\_s:\\{ * \\}\to S\\}\$$ where \$$f_s(*)=s\$$. Each \$$b_s\$$ such that \$$h\circ b_s=f_s\$$ must pick out an element in \$$T\$$, since the images of each \$$f_s\$$ are distinct, the images of each \$$b_s\$$ must also be distinct, so \$$\lvert T\rvert\geq\lvert S\rvert\$$ and \$$h\$$ must be surjective. Uniqueness of \$$b_s\$$ forces \$$h\$$ to be injective, so \$$h\$$ is an isomorphism and \$$T\cong S\$$. This shows that following defintion 3.87 agrees with following theorem 3.90.