Let $\Omega = \{ (x, y, z): z = x + 2y + 3 \}$ and $S = \{ (x, y, z): x^2 + y^2 + z^2 = 1 \}$. Let $\bar{a} = (x_n, y_n, z_n)$ a sequence in $\mathbb{R^3}$ such that $\lim_{n \to \infty} \bar{a_n} = \bar{a_0}$, $\bar{a_0} = (x_0, y_0, z_0) \in \mathbb{R^3}$.
Prove that $\forall n \in \mathbb{N} (\bar{a_n} \in \Omega \Longrightarrow \bar{a_0} \in \Omega) \wedge (\bar{a_n} \in S \Longrightarrow \bar{a_0} \in S)$.
I don't know the correct answer, so please let me know if I'm wrong or if you have better answer.
We know that $$\lim_{n \to \infty} (x_n, y_n, z_n) = (x_0, y_0, z_0) \Longleftrightarrow \lim_{n \to \infty} x_n = x_0 \quad \wedge \quad \lim_{n \to \infty} y_n = y_0 \quad \wedge \quad \lim_{n \to \infty} z_n = z_0$$
Proof of $$\forall n \in \mathbb{N} (\bar{a_n} \in \Omega \Longrightarrow \bar{a_0} \in \Omega)$$
Per hypothesis, since $\forall n\in \mathbb{N}, ~ \bar{a_n} = (x_n, y_n, z_n)\in \Omega$, $\bar{a_n}$ is of the form $z_n = x_n + 2y_n + 3$. Taking the limit on both sides of the equation $$\lim_{n \to \infty} z_n = \lim_{n \to \infty} (x_n + 2y_n + 3) = \lim_{n \to \infty} x_n + 2\lim_{n \to \infty}y_n + 3 \Longrightarrow z_0 = x_0 + 2y_0 + 3 \Longrightarrow \bar{a_0} = (x_0, y_0, z_0) \in \Omega$$
Proof of $$\forall n \in \mathbb{N} (\bar{a_n} \in S \Longrightarrow \bar{a_0} \in S)$$
Per hypothesis, since $\forall n\in \mathbb{N}, ~ \bar{a_n} = (x_n, y_n, z_n)\in S$, $\bar{a_n}$ is of the form $x_n^2 + y_n^2 + z_n^2 = 1$. Taking the limit on both sides of the equation $$\lim_{n \to \infty} (x_n^2 + y_n^2 + z_n^2) = \lim_{n \to \infty} x_n^2 + \lim_{n \to \infty} y_n^2 + \lim_{n \to \infty} z_n^2 = 1$$ Since $$\lim_{n \to \infty} x_n^2 = \lim_{n \to \infty} x_n\cdot x_n = \lim_{n \to \infty} x_n\cdot \lim_{n \to \infty} x_n = x_0 \cdot x_0 = x_0^2$$ we are having that $$x_0^2 + y_0^2 + z_0^2 = 1 \Longrightarrow \bar{a_0} = (x_0, y_0, z_0) \in S$$