[proofplan] We prove the sequential conclusion directly from the $\varepsilon$-$\delta$ definition of continuity at the point $a$. Given an arbitrary tolerance $\varepsilon > 0$, continuity supplies a radius $\delta > 0$ on which $f(x)$ stays within $\varepsilon$ of $f(a)$ for all $x \in E$. The convergence $x_n \to a$ then ensures that all sufficiently late terms $x_n$ lie inside that relative $\delta$-neighbourhood of $a$, so the corresponding values $f(x_n)$ lie within $\varepsilon$ of $f(a)$. [/proofplan] [step:Choose the continuity radius for an arbitrary output tolerance] Let $\varepsilon > 0$ be arbitrary. Since $f: E \to \mathbb{R}$ is continuous at $a \in E$, there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} [guided] We want to prove that the sequence of [real numbers](/page/Real%20Numbers) $(f(x_n))_{n=1}^{\infty}$ converges to $f(a)$. By the definition of convergence of a real sequence, we must start with an arbitrary tolerance $\varepsilon > 0$ and eventually force the inequality \begin{align*} |f(x_n)-f(a)| < \varepsilon. \end{align*} The hypothesis that $f$ is continuous at $a$ is precisely the tool that converts closeness of the input variable $x$ to $a$ into closeness of the output value $f(x)$ to $f(a)$. Because the domain of $f$ is the subset $E \subset \mathbb{R}$, this is relative continuity on $E$: there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} The number $\delta$ depends on the chosen $\varepsilon$ and on the function $f$ at the point $a$. Its role is to describe how close the sequence terms must be to $a$ before continuity guarantees the desired bound on their images under $f$. [/guided] [/step] [step:Use convergence of the sequence to enter the continuity neighbourhood] Let $\mathbb{N}=\{1,2,3,\dots\}$ denote the set of positive integers. Since $(x_n)_{n=1}^{\infty}$ is a sequence in $E$ and $x_n \to a$, the definition of sequence convergence applied to the number $\delta > 0$ gives an integer $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ with $n \ge N$, \begin{align*} |x_n-a| < \delta. \end{align*} For such $n$, we also have $x_n \in E$ because the sequence is a sequence in $E$. Therefore the implication supplied by continuity applies with $x = x_n$, and hence \begin{align*} |f(x_n)-f(a)| < \varepsilon. \end{align*} [/step] [step:Conclude convergence of the image sequence] We have shown that for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for every $n \ge N$, \begin{align*} |f(x_n)-f(a)| < \varepsilon. \end{align*} By the definition of convergence of a real sequence, this means that $(f(x_n))_{n=1}^{\infty}$ converges to $f(a)$. Equivalently, \begin{align*} \lim_{n \to \infty} f(x_n) = f(a). \end{align*} [/step]