> [!NOTE] Theorem (Algebra of Continuous Functions Over Euclidean Spaces) > Let $U\subset \mathbb{R}^n$, $p\in U$ and $f,g: U \to \mathbb{R}^k$ be [[Continuous maps|continuous]] at $p$. Then > 1. for all $a,b\in \mathbb{R}$, $af+bg$ is also cts at $p$. > 2. $fg$ is cts at $p$ > 3. let $h:U\to \mathbb{R}$ such that $h(p) \neq 0$, then $f/h$ is continuous at $p$. ###### Proof Applying [[Equivalence of Continuous and Sequentially Continuous Functions Over Euclidean Spaces|sequential continuity]], $f,g$ are continuous at $p$ iff for every sequence $(x_{j})$ in $U$ such that $x_{j}\to p$, we have $f(x_{j})\to f(p)$. Hence applying [[Algebra of Limits of Convergent Sequences]] yields $(1)$ and $(2)$. Now if $h(p) \neq 0$, applying $\varepsilon$-$\delta$ definition of continuity, there exists $\delta>0$ such that $\lVert x-p \rVert< \delta \implies \lvert h(x) - h(p) \rvert<\lvert h(p) \rvert \implies h(x)\neq 0$. Using quotient rule for limits, we can choose a sequence in $U$ such that $\lVert x_{j}-p \rVert< \delta$ and $x_{j}\to p$ then $\lim_{ j \to \infty }f(x_{j})/h(x_{j})=f(p)/h(p)$ since $h(x_{j})\neq 0$.