$L^2\cap C^\infty$ Hodge/Helmholtz decomposition on $\Bbb R^n$

I believe this is the part of the proof in question:

So far we have concluded the proof for $v \in C_{0}^{\infty}(\mathbb{R}^{N})$. To do it for a general $v \in L^{2}(\mathbb{R}^{N}) \cap C^{\infty}(\mathbb{R}^{N})$, we take $\rho \in C_{0}^{\infty}(\mathbb{R}), \rho \equiv 1$ for $|x| \leq 1, \rho \equiv 0$ for $|x| \geq 2$ and define $v_{n}(x) \equiv v(x) \rho(|x| / n)$. Then $v_{n} \in C_{0}^{\infty}(\mathbb{R}^{N})$, and $v_{n} \rightarrow v$ in $L^{2}$, so passing to the limit as $n \nearrow \infty$ we conclude the proof.

1. First, I'll summarize the steps before this. What we have is a triple sequence of smooth functions $(v_n,\nabla q_n, w_n)$ $$ v_n = v\rho(\bullet/n), \ v_n = w_n + \nabla q_n,$$ with the derivative estimates for all $\beta$, $$ \|D^\beta v_n\|_0^2 = \|D^\beta w_n\|_0^2 + \|\nabla D^\beta q_n\|_0^2$$ By construction $v_n$ converges to $v$ in every $H^s$. Also, we can set $q_n$ as the unique $H^1$ solution to the elliptic equation $$ \Delta q_n = \nabla\cdot v_n.$$ $\nabla q_n$ converges strongly in $L^2$ because it is given by a bounded Fourier multiplier on $v_n$, $$ \nabla q_n = \nabla \Delta^{-1}(\nabla\cdot v_n), \quad \widehat{\nabla q_n} = c\xi|\xi|^{-2}(\xi\cdot \hat v_n)$$ (or by the more direct proof in the book.) Hence $ w_n = v_n - \nabla q_n $ converges strongly in $L^2$.

2. Let $k\in\mathbb N$ be arbitrary. To conclude a similar result in each $H^k$, we differentiate $\alpha$ times, $|\alpha|=k$, $$ \underbrace{D^\alpha v_n}_{=V_n} = \underbrace{D^\alpha w_n}_{=W_n} + \nabla \underbrace{D^\alpha q_n}_{=Q_n} $$ and repeat the above argument for the triple sequence of smooth functions $(V_n, \nabla Q_n, W_n)$ to see that they also converge in $L^2$. Since $k$ is arbitrary (and fussing with uniqueness of limits) we have shown that the three $L^2$ limits $(v,\nabla q,w)$ are in fact elements of $H^k$ for all $k\ge 0$. By Sobolev embedding, they are $C^\infty \cap L^2$.