Suppose that instead of rectangular coordinates $x $ and $y $, we introduce curvilinear coordinates $u $ and $v $, where $$x = x(u,v) \space \text { and } \space y = y (u,v) $$ And $$ \begin{vmatrix} x_u & x_v \\ y_u & y_v \\ \end{vmatrix} \ne 0$$ Then the curve given by $y=y (x) $ corresponds to the curve given by some equation $v = v (u) $ in the $uv $-plane. The functional $$J [y] = \int_a^b F (x,y,y')dx $$ becomes $$J_1 [v] = \int_{a_1}^{b_1} F [x (u,v),y (u,v), \frac {y_u + y_v v'}{x_u+x_v v'}] (x_u+x_v v')du = \int_{a_1}^{b_1} F_1[u,v,v']du$$
Let $\Delta \sigma$ denote the area bounded by the curves $y = y(x)$ and $y = y(x)+h(x)$ and $\Delta \sigma_1$ denote the area bounded between the curves $v = v(u)$ and $v = v(u)+\eta(u)$
As my book states, "as $\Delta \sigma, \Delta \sigma_1 \to 0$, the ratio $\Delta \sigma / \Delta \sigma_1 $ approaches the Jacobian, which is non zero by hypothesis. Thus $$\lim_{\Delta \sigma \to 0} \frac{J[y+h]-J[y]}{\Delta \sigma} = 0$$ Implies $$\lim_{\Delta \sigma_1 \to 0} \frac{J[v+\eta]-J[v]}{\Delta \sigma_1} = 0$$
So that $v$ is extremal if and only if $y$ is extremal."
I don't understand how the ratio of areas approaching a non-zero number in the limit that they go to zero implies that
$$\lim_{\Delta \sigma_1 \to 0} \frac{J[v+\eta]-J[v]}{\Delta \sigma_1} = 0$$ If $$\lim_{\Delta \sigma \to 0} \frac{J[y+h]-J[y]}{\Delta \sigma} = 0$$
If anyone has notational questions, let me know and I will clarify
Personally I find the derivation slightly clearer when variations are written as $w(x) = \hat{w} (x) + \epsilon \beta(x)$ and the limits are performed with respect to the variation parameter $\epsilon$, not $\Delta \sigma$, for example as done in http://www.physics.usu.edu/Wheeler/ClassicalMechanics/CMCoordinateinvarianceofEulerLagrange.pdf.
Anyhow, I believe the point made by your textbook could be explained as follows.
If $$ \frac{ \Delta \sigma}{\Delta_1 \sigma } \to I $$ ($I$ standing for the Jacobian) as $\Delta \sigma, \Delta_1 \sigma \to 0$, one could write to first order $$ lim_{\Delta \sigma \to 0} \frac{J[y + h] – J[y]}{\Delta \sigma} = lim_{\Delta_1 \sigma \to 0} \frac{J[y + h] – J[y]}{I \Delta_1 \sigma} = \frac{1}{I} lim_{\Delta_1 \sigma \to 0} \frac{J[y + h] – J[y]}{\Delta_1 \sigma} $$ The action at the numerator is invariant upon coordinate changes, so the fact that $I \neq 0$ and that the first limit equals $0$ implies the conclusion.