> [!NOTE] Theorem > Let $I$ be an [[Open Real Interval|open interval]]. Let $c\in I.$ Let $f:I\to \mathbb{R}$ be a [[Real Function|real function]]. Then $f$ is [[Continuous Real Function|continuous]] at $c$ iff the [[Limit of Real Function at a Point|limit]] of $f(x)$ as $x$ tends to $c$ is $f(c)$: that is, $\lim_{ x \to c }f(x)=f(c) $ **Proof**: Suppose $f$ is continuous at $c.$ Then for all $\varepsilon>0,$ there exists $\delta>0$ such that for all $x\in I,$ $0<|x-c|<\delta \implies |f(x)-f(c)|<\varepsilon.\tag{1}$that is, $\lim_{ x \to c }f(x)=f(c).$ Conversely, suppose $\lim_{ x \to c }f(x)=f(c).$ Then for all $\varepsilon>0,$ there exists $\delta>0$ such that for all $x\in I,$ $(1)$ is true. Now suppose $x\in I$ such that $|x-c|<\delta.$ Then either $x \neq c$ or $x=c.$ In the first case, $0<|x-c|<\delta$ so $|f(x)-f(c)|<\varepsilon.$ In the latter case, $f(x)-f(c)=f(c)-f(c)=0<\varepsilon.$ So indeed for all $x\in I,$ $|x-c|<\delta \implies |f(x)-f(c)|<\varepsilon$that is, $f$ is continuous at $c.$