Does the functor that preserves limit always have a left adjoint? If $F: Sets \to Sets$ is a functor that preserves limits, is it true that $F$ always has a left adjoint?
 A: Yes. This is a corollary of the special adjoint functor theorem. 
A: Every continuous functor $F : \mathsf{Set}\to \mathsf{Set}$ is isomorphic to $\hom(U,-)$ for some set $U$ and hence right adjoint to $- \times U$. Qiaochu has already mentioned that this follows from the Special Adjoint Functor Theorem. In the special case of $\mathsf{Set}$, one can make this more explicit as follows:
Let $F : \mathsf{Set} \to \mathsf{Set}$ be a continuous functor. Let $2=\{0,1\}$. 
Let $S=F(2)$. Consider $2^S$. Since $F$ is continuous, we have $F(2^S) \cong F(2)^S = S^S$. There is a canonical element in $S^S$, the identity map $S \to S$. We choose a preimage $u \in F(2^S)$. We consider subsets $T \subseteq 2^S$ such that $u$ lies in the image of the induced injective map $F(T) \hookrightarrow F(2^S)$, let us just write $u \in F(T)$. These sets are closed under intersections (since $F$ is continuous). Hence there is a smallest subset $U \subseteq 2^S$ such that $u \in F(U)$. This is our $U$.
I will not prove $F \cong \hom(U,-)$ (the proof can be found here), but explain why the construction really gives $U$ when applied to $F=\hom(U,-)$: We have $S=\hom(U,2) \cong P(U)$, the power set of $U$. Then, $2^S \cong P(P(U))$. We have $u : U \to P(P(U))$, $x \mapsto \{V \subseteq U: x \in V\}$. The smallest subset of $P(P(U))$ over which $u$ factors is the image of $u$, which identifies with $U$.
