A question on large-N limit?

Question

Let's take $SU(N)$ for an example. The Lagrangian is $$\mathcal{L}=-\frac{1}{4g_{YM}^2}F_{\mu\nu}F^{\mu\nu}.$$ We can define the t'Hooft coupling as $$\lambda=g_{YM}^2N.$$ Then the large-$N$ limit or the t'Hooft limit is: $$N\rightarrow\infty,\ \text{but with}\ \lambda\ \text{fixed}.$$ I can understand why such a strategy lead us to the topological $1/N$ expansion. But a very basic question emerges. In the large-$N$ limit, we shall have $g_{YM}\rightarrow 0$, then why dont we simply do the perturbative expansion of $g_{YM}$? Further, if in the large-N limit we always have $g_{YM}\rightarrow 0$, then does it mean that we can only handle weak coupling situation in the large-$N$ expansion?

I think there is something I missed here. Since if the large-$N$ limit can only handle the weak coupling situation, then it will be useless. In contrast, it seems to be very useful in the literature, especially when we talk about strong coupling phenomena such as the colour confinement.

Answer 1

It is true that in the 't Hooft limit the individual contributions of the Feynman diagrams at order $n$ are going down by factors of $g^{2n}$. But remember that the number of Feynman diagrams at that order is scaling as $N^n$ (due to the growing number of color indices). Hence, the effective contributions scale according to the 't Hooft parameter $g^2N$.