the couplings $\lambda$ and $\lambda_6$ are both exactly marginal in this limit.

At finite but large $N$ the beta function does not vanish. Without the coupling $\lambda$, the beta function is positive so the theory with $\lambda_6 > 0$ is trivial (IR-free). However, when $\lambda \neq 0$ and for large $N \geq 10$, we find from two lines of non-trivial fixed points of the two-loop beta function, $$\begin{aligned}
  \lambda_{6}^\pm(\lambda) &=
  \frac{\left(20 N - 20 \pm \sqrt{1852-2054 N+202 N^2} \right) \lambda^2}
  {44+6 N} {\,.}\end{aligned}$$ The line $\lambda_6^+(\lambda)$ is IR-stable, while $\lambda_6^-(\lambda)$ is UV-stable – see Figure 1. Note that since $\beta_\lambda=0$, the renormalization group flow is always in the $\lambda_6$ direction.

\[fig:RG\] ![The space of coupling constants and the renormalization group flow towards the IR for large but finite $N$, based on the two-loop beta function. There are two lines of fixed points. Since $\beta_{\lambda}=0$, the flow lines are all in the $\lambda_6$ direction.](Beta6.png "fig:"){width="60.00000%"}

The Large $N$ Beta Function $\beta_{\lambda_6}$ to All Orders {#allorders}
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In this section we argue that $\beta_{\lambda_6}(\lambda,\lambda_6) = O(1/N)$ to all orders in perturbation theory, generalizing our explicit two-loop computation of the previous subsection. We could not find a direct argument for this, so instead 