> [!NOTE] Theorem (Limit of Real Function by Convergent Real Sequences) >Let $f:I\to \mathbb{R}$ be a [[Real Function|real function]]. Let $c,L\in \mathbb{R}.$ Then [[Limit of Real Function at a Point|limit]] of $f$ at $c$ equals $L$ iff for all [[Real sequences|sequence of real numbers]] $(x_{n})_{n\geq 0}$ in $I\setminus\{ c \}$ that [[Convergence|converge]] to $c$ we have $f(x_{n})\to L$ as $n\to \infty.$ **Proof**: ($\implies$) Since $\lim_{ x \to c } f(x) = L,$ for all $\varepsilon>0$ there exists $\delta>0$ such that for all $x\in I,$ $0<|x-c|<\delta \implies |f(x)-L|<\epsilon.$Let $x_{n}\in I \setminus \{ c \}$ such that $x_{n}\to c.$ Then there exists $N\in \mathbb{N}$ such that for all $n \geq N,$ $ 0<|x_{n}-c|<\delta$So for all $n \geq N,$ $|f(x_{n})-L|<\varepsilon$thus $f(x_{n})\to L$ as $n\to \infty.$ ($\impliedby$) Conversely, suppose $\lim_{ x \to c }f(x)\neq L$. Then there exists $\varepsilon>0$ such that for all $\delta>0,$ there exists $x\in I$ such that $0<|x-c|<\delta \quad \land \quad |f(x)-L|\geq \varepsilon.$Choose $(x_{n})_{n \geq 0}\in I$ such that $|x_{n}-c|< \frac{1}{n}$ and $|f(x_{n})-L|\geq \varepsilon.$ Then $x_{n}\to c$ but $f(x_{n})\not\to c$ as $n\to \infty.$