The idea of the question is whether it is ever possible that $L$ is so nice in the sense that $\{L\_\alpha\}$ does not incorrectly "guess" a bigger inaccessible than $L$ really has, as long as $L\_\alpha \models ZFC + V = L$. 


More precisely, is it ever possible that $L \models "\kappa$ is the largest inaccessible" and for all $\alpha > \kappa$ with $L\_\alpha \models ZFC + V = L$ (or perhaps large finite fragments), $L\_\alpha \models "\kappa$ is largest inaccessible". 


Are there assumption that one can put on $V$ so the it's $L$ has the above property? 


By reflection, there are unboundedly many $\alpha$ such that $L\_\alpha \models "\kappa$ is largest inaccessible". However, it appear that potentially there are many $\alpha$'s which could "temporarily" think some ordinal greater than $\kappa$ is an inaccessible cardinal. 


Thanks for any information or clarification.



Property of $L$ Relating to Reflection
The answer to the "More precisely ..." question is that yes, it may be the case that $L$ has a single inaccessible cardinal $\kappa$ and that for no larger $\beta >\kappa$ is there any $\alpha$ with $L\_\alpha\models$``$\beta$ is inaccessible.'' For example if $\kappa$ is the first inaccessible in $L$ (if such exists) and there is a least $\delta$ with $L\_\delta\models ZFC$ (if such exists) then $L\_\delta$ is a model for this property: namely for *every* $\alpha \in (\kappa,\delta)$ we have $L\_\alpha\models \kappa $ is the largest inaccessible.


The restrictive assumption to put on $V$ to ensure that $L$ has this property is just to rule it out by fiat: that `There is no transitive model of $ZFC + \exists \kappa(\kappa $ is inaccessible)''. 


