Coupling Constant under Infrared Limit

Question

In the paper "More Abelian Dualities in 2+1 Dimensions",

https://arxiv.org/abs/1609.04012

on page 4, it is said that the theory

$$S[\phi;a]=\int d^{3}x\left\{\frac{-1}{4e^{2}}f_{\mu\nu}f^{\mu\nu}+|\left(\partial_{\mu}-ia_{\mu}\right)\phi|^{2}-\alpha|\phi|^{4}\right\},$$

where $f_{\mu\nu}=\partial_{\mu}a_{\nu}-\partial_{\nu}a_{\mu}$, in the infrared limit ($e\rightarrow\infty$ $\alpha\rightarrow\infty$), has no Maxwell term. i.e. $\frac{1}{4e^{2}}da\wedge\star da=0$.

My understanding is the following. 

1. Under a change of variables: $x=x^{\prime}/b$, where $0<b<1$, one has

   \begin{align}
S[\phi;a]
& = \int d^{3}x\left\{\frac{-1}{4e^{2}}f_{\mu\nu}f^{\mu\nu}+|\left(\partial_{\mu}-ia_{\mu}\right)\phi|^{2}-\alpha|\phi|^{4}\right\}
\\ & =\int d^{3}x^{\prime}\left\{\frac{-1}{4e^{2}}b^{-1}(\partial^{\prime}_{\mu}a_{\nu}-\partial^{\prime}_{\nu}a_{\mu})(\partial^{\prime\mu}a^{\nu}-\partial^{\prime\nu}a^{\mu})
\right. \\ & \qquad \qquad \qquad \left. +b^{-1}\left(\partial^{\prime}_{\mu}\phi^{\ast}\right)\left(\partial^{\prime\mu}\phi\right)+ib^{-2}a_{\mu}\phi^{\ast}\left(\partial^{\prime\mu}\phi\right)
\right. \\ & \qquad \qquad \qquad \left. 
-ib^{-2}a^{\mu}\left(\partial^{\prime}_{\mu}\phi^{\ast}\right)\phi+b^{-3}a_{\mu}a^{\mu}\phi^{\ast}\phi-b^{-3}\alpha |\phi|^{4}\right\},
\end{align}

   where $\partial^{\prime}_{\mu}=\partial/\partial x^{\prime\mu}$. 

2. Defining new fields $\phi^{\prime}=b^{-1/2}\phi$, $a^{\prime}_{\mu}=b^{-1}a_{\mu}$ and $f^{\prime}_{\mu\nu}=\partial^{\prime}_{\mu}a^{\prime}_{\nu}-\partial^{\prime}_{\nu}a^{\prime}_{\mu}$, one has
   \begin{align}
S[\phi;a]& =\int d^{3}x^{\prime}\left\{\frac{-1}{4e^{2}}bf^{\prime}_{\mu\nu}f^{\prime\mu\nu}+\left(\partial^{\prime}_{\mu}\phi^{\prime\ast}\right)\left(\partial^{\prime\mu}\phi^{\prime}\right)+ia^{\prime}_{\mu}\phi^{\prime\ast}\left(\partial^{\prime\mu}\phi^{\prime}\right)
\right. \\ & \qquad \qquad \qquad \left. 
-ia^{\prime\mu} \left(\partial^{\prime}_{\mu}\phi^{\prime\ast} \right)\phi^{\prime}+a^{\prime}_{\mu}a^{\prime\mu}\phi^{\prime\ast}\phi^{\prime}-b^{-1}\alpha |\phi^{\prime}|^{4}\right\}
\end{align}

3. Defining new couplings $1/e^{\prime 2}=b/e^{2}$, and $\alpha^{\prime}=b^{-1}\alpha$, one has 

   $$S[\phi;a]=S[\phi^{\prime};a^{\prime}]=\int d^{3}x^{\prime}\left\{\frac{-1}{4e^{\prime 2}}f^{\prime}_{\mu\nu}f^{\prime\mu\nu}+|\left(\partial^{\prime}_{\mu}-ia^{\prime}_{\mu}\right)\phi^{\prime}|^{2}-\alpha^{\prime}|\phi^{\prime}|^{4}\right\}$$

   Under the change of variables $x=x^{\prime}/b$, the infrared limit is $b\rightarrow 0^{+}$. This is equivalent to $e^{\prime}\rightarrow\infty$ and $\alpha^{\prime}\rightarrow\infty$.

Is that correct? Do I need to compute the corresponding beta functions?