**Theorem** If $f,g:E \to \mathbb{R}$ are both [[Continuous Real Function|continuous]] at $c \in E$, then: 1. $f+g$ is continuous at $c$ 2. $fg$ *(not $f \circ g$)* is continuous at $c$ 3. If $g(c) \neq 0$ then $\frac{f}{g}$ is defined on $E \cap (c-\delta,c+\delta)$ for some $\delta>0$, and $\frac{f}{g}$ is continuous at $c$. **Proof** 1. Using [[Algebra of Limits of Convergent Sequences|sum rule for limits]], $f(x_{n})+g(x_{n}) \to f(c)+g(c) = (f+g)(c)$ 2. Using product rule for limits, $f(x_{n})g(x_{n}) \to f(c)g(c) = (fg)(c)$ 3. Using [[Preservation of Inequalities by continuous real functions]], since $g(c) \neq 0$ then $g(x) \neq 0$ on $E' := (c-\delta,c+\delta) \cap E$ for some $\delta>0$. So $\frac{f}{g}$ is defined on $E'$. Taking a sequence $(x_{n})\in E'$ with $x_{n} \to c$, we have using quotient rule for limits that $(f/g)(x) =\frac{f(x_{n})}{g(x_{n})}\to \frac{f(c)}{g(c)} = (f/g)(c). $ **Remarks** Note that we can deduce that any [[Ring of Polynomial Forms|univariate polynomial]] is everywhere continuous. Also any rational function of polynomials is continuous provided the denominator is non-zero.