$V_\lambda$ models I have two question about $V_\lambda$ models:


*

*I know that, if $\kappa$ is strongly inaccessible then
$$
V_\kappa\models ZFC
$$
But, the converse statement is true? i.e. if $\lambda$ is a limit ordinal such that $V_\lambda\models ZFC$, then $\lambda$ must be strongly inaccessible cardinal? 

*is there a limit ordinal $\lambda$ such that $V_\lambda\nvDash(\text{replacement axiom})$?


I think that the first question is not true, but I can't find a counterexample. For the second question I don't have idea. 
Can someone help me? Thanks for advance.
 A: For $(1)$, the answer is no; if $V_\kappa\models$ZFC we say $\kappa$ is a worldly cardinal, and this is a strictly weaker condition than strong inaccessibility. 
In particular, it's easy to see that (under reasonable hypotheses) we can have a singular worldly cardinal. Working in an appropriate model of ZFC, for $\alpha$ an ordinal let $f(\alpha)$ be the least height of a transitive elementary submodel containing $V_\alpha$ as an element. Then $\lambda=\sup\{0, f(0), f(f(0)), ...\}$ is worldly, since $V_\lambda$ is the union of a chain of elementary submodels. However, we have $cof(\lambda)=\omega$, hence $\lambda$ is not inaccessible.

For $(2)$, the answer is yes: as long as $\lambda$ is "easily seen" to be singular, then $V_\lambda$ won't satisfy Replacement. For example, $V_{\omega^2}$ doesn't satisfy Replacement since the map $$\omega\rightarrow\omega^2: n\mapsto \omega\cdot n$$ is definable inside $V_{\omega^2}$.
(Note that since Replacement is a scheme rather than a single axiom, there could be some interesting distinctions to be made in terms of how much Replacement a given level of the cumulative hierarchy - or indeed any "reasonable" transitive set - satisfies. If you're interested in this sort of thing you might like thinking about admissible sets.)
