Proof that the Gaussian Curvature is a Ratio of Areas Let $S \subset \mathbb{R}^3 $ be a smooth surface, and let $S^"$ be the unit sphere, and let $n: S \to S^2$ be a given Gauss map.
I want to prove that the Gaussian curvature $K(p)$ at a point $p \in S$ is the limit of the ratio of areas we get when a neighbourhood $U$ shrinks around $p$ :
$$ K(p) = \lim_{U \to p}\frac{\text{Area}(n(U))}{\text{Area}(U)} $$
I have shown already that
$$\lim_{U \to p} \frac{\text{Area}(n(U))}{\text{Area}(U)} \le K(p) $$
letting $U = F(V)$ for some local parameterization $F$ from $V \subset \mathbb{R}^2$ to $U \subset S$, and using my definition:
$$
\text{Area}(U) := \int_V ||\partial_x F \times \partial_y F || dx dy.
$$
and the triangle inequality.
Is a lower bound obvious? or do we need to use the (reverse) triangle inequality or something?
 A: Fist of all letme recast your original formulation:
\begin{equation}
|K(p)| = \lim_{U \to p}\frac{\text{Area}(n(U))}{\text{Area}(U)}
\end{equation}
I added an absolute value since the ratio of areas is non negative. Now let $X:V \rightarrow \mathbb{R}^{3}$ be a regular parametrization of a neighborhood of $p=X(u_0,v_0)$, where V is an open subset of $\mathbb{R}^2$ . Since X(V) is an open set, $U_ε \subset X(V)$ for all ε. Since we assumed that K(p) is non null, when considering the continuity of the Gaussian curvature function $K:V \rightarrow \mathbb{R}$, we can conclude that the Gaussian curvature does not change sign for ε chosen small enough. We can define the preimage $V_{\epsilon}=X^{-1}(U_{\epsilon})$. Since X is a regular parametrization, $(u_0,v_0)$ is the unique point in all $V_{\epsilon}$ for all positive $\epsilon$.Then by using the formula of the surface are, it is possible to write:
\begin{equation}
Area(U_{\epsilon})=\int \int_{V_{\epsilon}} ||X_u \times X_v|| du dv
\end{equation}
In a similar way on the unit sphere we have
\begin{equation}
Area(n(U_{\epsilon}))=\int \int_{V_{\epsilon}} ||N_u \times N_v|| du dv = \int \int_{V_{\epsilon}} |K(u,v)| ||X_u \times X_v|| du dv
\end{equation}
By using the Mean Theorem or double integrals we can write
\begin{equation}
 \int \int_{V_{\epsilon}}||X_u \times X_v|| du dv = ||X_u(u_{\epsilon},v_{\epsilon}) \times X_v(u_{\epsilon},v_{\epsilon})||
\end{equation}
And
\begin{equation}
 \int \int_{V_{\epsilon}}||N_u \times N_v|| du dv = |K(u^{'}_{\epsilon},v^{'}_{\epsilon})|||X_u(u^{'}_{\epsilon},v^{'}_{\epsilon}) \times X_v(u^{'}_{\epsilon},v^{'}_{\epsilon})||
\end{equation}
Thus, since when $\epsilon \rightarrow 0$ both $(u_{\epsilon},v_{\epsilon})$ and $(u^{'}_{\epsilon},v^{'}_{\epsilon})$ tend to $(u_0, v_0)$ we have:
\begin{equation}
\lim_{U \to p} \frac{\text{Area}(n(U))}{\text{Area}(U)} = \lim_{\epsilon \to 0} \frac{\text{Area}(n(U_{epsilon}))}{\text{Area}(U_{epsilon})}= \lim_{\epsilon \to 0} \frac{|K(u_{0},v_{0})|||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}{||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}=K(p)
\end{equation}
