The parameters $g^2$ and $N$ are independent of each other so it is meaningless to ask whether $g^2$ goes like $1/N$ for large or small coupling "in general". In general, it may go but it doesn't have to go.
But if it does go, i.e. if $g^2 N$ is kept fixed, then one may say new interesting things. If one introduces a new symbol $\lambda = g^2 N$ for the 't Hooft coupling, the condition simply says that $\lambda$ is kept fixed – which is natural, especially in the dual stringy interpretation of the same physics.
I need to dedicate a special paragraph to an error in your question which is probably not just a typo, due to the complicated work you had to go through to write the fractions. What is kept fixed in 't Hooft's limit isn't $g^2/N$; it is $g^2 N$. It's the product, not the ratio! So $g^2$ indeed goes like $1/N$ and not $N$ if $N$ is sent to infinity.