if $L$ satisfy the following:
$L$ is an uncountable cardinal
$L$ is a limit cardinal, i.e. $L$ is not a successor cardinal, i.e. there is no cardinal $\kappa$ such that $L$ is the least cardinal strictly bigger than $\kappa$.
$L$ is regular
Does that mean that $L$ must be inaccessible?
Uncountable regular limit cardinals are called weakly inaccessible. For a weakly inaccessible $\kappa$ to be inaccessible it also needs to be a strong limit, which means $2^{\lambda} < \kappa$ for all $\lambda < \kappa.$
(Note some references use the term "strongly inaccessible", rather than just "inaccessible", to contrast with the weak notion.)
It is consistent that every weakly inaccessible cardinal is inaccessible: If we assume GCH, then every limit is a strong limit, since $\lambda < \kappa$ implies $2^\lambda = \lambda^+ < \kappa.$
However, it is also consistent that there are weak inaccessibles that are not inaccessible. For instance, we can use Cohen forcing to move the continuum larger than the smallest weakly inaccessible $\kappa$. Cohen forcing preserves cardinals and regularity, so it preserves weak inaccessibility. Thus in the forcing extension, $\kappa$ is weakly inaccessible, but it is clearly not a strong limit since $2^{\aleph_0}$ is larger.