This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.
Choose, for each infinite cardinal $\alpha$, a field $F_\alpha$ of that cardinality (say, for definiteness, the characteristic zero algebraically closed field of transcendence degree $\alpha$ over $\mathbb{Q}$), and put $A_\alpha = \mathbb{Z} \times F_\alpha$. Each non-trivial quotient ring (corresponding to a regular epi) of $A_\alpha$ either contains a copy of $F_\alpha$, or is a quotient ring of $\mathbb{Z}$ (possibly $\mathbb{Z}$ itself).
This has the following consequence: for any (commutative) ring $R$ of cardinality less than $\alpha$, there is exactly one map $f: A_\alpha \to R$. For if we have a (regular epi)-mono factorization $A_\alpha \to Q \to R$ where $Q \to R$ is monic, then the possibility where $Q$ contains a copy of $F_\alpha$ is ruled out, hence the factorization must take the form
$$A_\alpha \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$
where $(n)$ is uniquely determined as the annihilator of the identity in $R$.
Now form the functor
$$G = \prod_{\alpha \in \text{Card}} \hom(A_\alpha, -): \textbf{CRing} \to \textbf{Set}$$
As soon as $\alpha \gt \text{Card}(R)$, we have that $\hom(A_\alpha, R)$ is a one-element set. Thus for each $R$, $G(R)$ is a set even though $G$ itself is a class-sized product. Being a product of continuous functors, $G$ is continuous. But $G$ cannot be representable (just by simple cardinality considerations; e.g., $G(A_\alpha)$ has size greater than $\alpha$, for any $\alpha$, since algebraically closed fields have lots of automorphisms).