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.
For any set $S$, let $\mathbb{Z}[S]$ be the free commutative ring on $S$, aka the polynomial algebra with coefficients in $\mathbb{Z}$ and indeterminates drawn from $S$. Let $A_S$ denote the localization of $\mathbb{Z}[S]$ where all elements except those belonging to $\mathbb{Z}$ are inverted. It follows that the only quotient rings (corresponding to regular epis) of $A_S$ are $\mathbb{Z}$ and its quotient rings.
This has the following consequence: for given $S$ and any (commutative) ring $R$, there is exactly one non-injective map $f: A_S \to R$. For, any such $f$ factors uniquely as
$$A_S \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$
where $(n)$ is uniquely determined as the annihilator of the identity in $R$.
Now, for each cardinal $\alpha$, let $S_\alpha$ be a set of that cardinality (for example, we could take $S_\alpha$ to be the von Neumann cardinal $\alpha$ itself, considered as a set), and form the functor
$$F = \prod_{\alpha \in \text{Card}} \hom(A_{S_\alpha}, -): \textbf{CRing} \to \textbf{Set}$$
The main thing to notice is that for any ring $R$, as soon as $\alpha \gt \text{Card}(R)$, we cannot have an injection $A_{S_\alpha} \to R$, in which case $\hom(A_{S_\alpha}, R)$ is a one-element set. Thus for each $R$, $F(R)$ is a set even though $F$ itself is a class-sized product. Being a product of continuous functors, $F$ is continuous. But $F$ cannot be representable (just by simple cardinality considerations; e.g., $F(A_{S_\alpha})$ has size at least $\alpha^\alpha$, for any $\alpha$).