Arguments to show that $\lim_{\iiint d\tau\to 0}\frac{\iint d\vec{\sigma}\times\vec{V}}{\iiint d\tau}=\nabla\times\vec{V}$

Question

So, I was solving the problem 1.10.6 from "George B Arfken, Hans J Weber - Mathematical Methods For Physicists- Sixth edition" and the problem was to show that:

$\lim_{\iiint d\tau\to 0}\frac{\iint d\vec{\sigma}\times\vec{V}}{\iiint d\tau}=\nabla\times\vec{V}$

And he says to choose the volume $\iiint\tau$ to be the volume differential $dxdydz$.

I think that maybe I got the solution, but I'm not sure if that's really correct, and if it is, I have two main points that brought me here:

1. The argument between "*** ***" don't see much convincing to me, I think that it should be something like this, but I don't know if that's strictly correct. So I would like to know if that's is really correct and, if it's, how could I improve this to sound and look better. In the argument between " *** *** " can I use the argument of the orientation of the surface ? Since there isn't a vector surface integral, I'm integrating the component $V_k$ which is a scalar function, so still make sense use the argument of the orientation of the surface for the signal of $dx_jdx_i$?
2. I saw in the book some proofs that take steps similar to the step (3)$\to$(4) where the author just get rid of the integrals, so I suppose that I can do this that way too in this specific situation (can I ?), but I'm not sure. And I don't really understand why I would be able to do so. So, I'm asking here if someone could explain it to me. I would be really grateful for that

EDIT: I think that maybe I got the idea of the passage between the step $(3)$ and $(4)$. We define the integrals as the limit of the $\Delta x_i$ going to zero from the Riemann sum, and then the $\Delta x_i$ becomes an infinitesimal element $dx_i$, but since we are evaluating the variations of $V_k$ at this limit, it doesn't make sense talk about integrals, then we just keep the integrand expanded in the Taylor series and the sums, but I'm not 100% sure of this argument, is that right?

Note. I edited the point number 1 because I think that the question wasn't clear enough, but still, the same question, can I argue that way?

Follow my resolution of the problem below:

I choose the surface to be an infinitesimal cube centered on the origin so its bounded by the planes $x = \frac{dx}{2},x = -\frac{dx}{2},y = \frac{dy}{2},y = -\frac{dy}{2},z = -\frac{dz}{2},z = -\frac{dz}{2}$, and it's volume is $dxdydz$.

then I define the infinitesimal area vector $d\vec{\sigma}$, and the vector field $\vec{V}$ to be:

$\begin{equation} d\vec{\sigma} = (dydz,dxdz,dxdy)\quad\text{and}\quad \vec{V} = (V_x,V_y,V_z)\end{equation}$

Then using the Levi-Civita symbol :

$d\vec{\sigma}\times\vec{V} = \epsilon_{ijk}dx_i dx_k V_k \ \ \ \ \ \ \ (1)$

with $dx,dy,dz,V_x,V_y,V_z$ being $dx_1,dx_2,dx_3,V_1,V_2,V_3$ respectively.

Now from (1) we get that:

$\iint d\vec{\sigma}\times\vec{V} = \iint \epsilon_{ijk}dx_i dx_k V_k = \epsilon_{ijk}\iint V_k dx_i dx_k \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

*** Now, we have an infinitesimal closed cube surface, and the directions of the six sides of the cube are defined to be outwards, so we have two sides of the cube defined by $dx_idx_k$ (one by $-dx_idx_k$ precisely), both over the $x_j$ axis one where $x_j = \frac{dx_j}{2}$ and the other where $x_j = -\frac{dx_j}{2}$. So with that in mind, we have to evaluate $V_k$ on both points in order to proceed with the problem, expanding $V_k$ till the first-order Taylor series ***:

$\iint d\vec{\sigma}\times\vec{V} = \epsilon_{ijk}\big[\iint \big(V_k -\frac{\partial V_k}{\partial x_j}\frac{dx_j}{2}\big) \big(-dx_i dx_k\big) + \iint \big(V_k +\frac{\partial V_k}{\partial x_j}\frac{dx_j}{2}\big)dx_i dx_k \big] \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

and from (3) we get that:

$\iint d\vec{\sigma}\times\vec{V} = \epsilon_{ijk} \frac{\partial V_k}{\partial x_j}dx_idx_jdx_k \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

dividing by $\iiint\tau = dx_idx_jdx_k$ we find that:

$\lim_{\iiint\tau\to 0}\frac{\iint d\vec{\sigma}\times\vec{V}}{\iiint d\tau}= \epsilon_{ijk} \frac{\partial V_k}{\partial x_j}=\nabla\times\vec{V}$

And we can note that if we expands $V_k$ to the high order terms from the Taylor series, those terms go to $0$ when $\iiint\tau\to 0$, because $dx_i\to 0,dx_j\to 0,dx_k\to 0$