Definition of a limit when there are no cones? A very pedantic question: If a category $\mathcal{C}$ admits no cones, then does it contains all limits vacuously? Or does the definition of a limit imply the existence of at least one limit?
 A: You should distinguish more carefully between a diagram which admits no cones and a category which admits no cones. As the other answer explains, it is only the empty category that admits no cones over any diagram, so this property is not very interesting.
Let’s consider the other property, then. For instance the identity functor of the discrete category on two objects admits no cone, and this implies it admits no limit, since a limit cone is in particular a cone. This is one of the three ways a diagram can lack a limit: it can have no cones at all, or it can no cone through which every other cone factors (this is called a weak limit cone) or it can have no weak limit cone, factorizations through which are always unique.
A: Every nonempty category admits at least one cone (not to say universal cone, but just a cone): a cone for any functor $X:\ast \to \mathscr{C}$ which picks out a fixed object $X$ of $\mathscr{C}$ and the identity morphism on $X$. To find a category $\mathscr{C}$ that admits no cones, you necessitate that the category be not nonempty.
The only category that has no cones is $\emptyset$. This category admits exactly one functor (the identity functor, which is the empty functor), and this has no cones because there do not exist any objects in $\emptyset$. In particular, a limit to $id_{\emptyset}$ is an object in $\emptyset$ so $\emptyset$ cannot admit any limits.
