Suppose the function \begin{equation} h(x,y) = \frac{\sin\sqrt{1-x^2-2y^2}}{\sqrt{1-x^2-2y^2}} \end{equation} is defined on $\left\{(x,y)\vert x^2 + 2y^2 < 1\right\}$. We need to prove that \begin{equation} \forall x_0 \in \mathbb{R}, \forall y_0 \in \mathbb{R}, {x_0}^2 + 2{y_{0}}^{2} = 1 \longrightarrow \lim_{(x,y)\to(x_0,y_0)}h(x,y) = 1 \end{equation}
In my proof, I assumed the existence of the following conclusions: \begin{equation} \lim_{u\to 0}\frac{\sin u}{u} = 1, \end{equation} \begin{equation} \forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0. \end{equation}
To strictly follow the rules of first order logic, I applied the fitch style. The proof is divided into two parts. In the first part, I deduced the following proposition:
\begin{equation}
\begin{aligned}
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0\right) \longrightarrow\\
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow
\forall \varepsilon > 0, \exists \delta > 0, \forall x, \forall y, \left(0 < \sqrt{(x-u)^2+(y-v)^2} < \delta \wedge x^2+2y^2 < 1\right) \longrightarrow \sqrt{1-x^2-2y^2} < \varepsilon\right)
\end{aligned}
\end{equation}
The formal proof is as follows:
In the second proof, the final conclusion is deduced. The formal proof is given below:
The formal proof of this seemingly simple conclusion is very tedious. However, without proper use of first order logic, there is no systematic way to examine a deduction. Recalling my first classes in undergraduate, I feel very upset about the type of proof I learned at that time. Every deduction was based on intuitive use of math rules, especially for quantification. I am wondering how to train myself to prove in English while have a picture of first order logic in mind.