Theorem: Assume that f is a function that is continuous at c, and assume that $\lim\limits_{x \to \infty}$$(s_{n})$ = c , where all the terms $s_{n}$ are in the domain of f. Then $\lim\limits_{x \to \infty}$$(f(s_{n}))$ = f(c).

How can this theorem be proven, using the precise limit definitions for functions and infinite sequences?

  • $\begingroup$ You are going to have to provide your definitions of continuity and limits, because as it happens, that is one potential way of defining “$f$ is continuous at $c$”. $\endgroup$ Nov 28, 2020 at 20:08

1 Answer 1



Let $f:X\to Y$ be a real-valued function over $X\subseteq\mathbb{R}$. Then $f$ is continuous at $x_{0}\in X$ if and only if $f(x_{n})\to f(x_{0})$ whenever $x_{n}\in X$ converges to $x_{0}$.


We shall prove the implication $(\Rightarrow)$ first.

Let us suppose that $f:X\to Y$ is continuous at $x_{0}\in X$. Then for every $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} > 0$ such that for every $x\in X$ one has that \begin{align*} |x - x_{0}| \leq \delta_{\varepsilon} \Rightarrow |f(x) - f(x_{0})| \leq \varepsilon \end{align*}

Moreover, since $x_{n}\to x_{0}$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ such that \begin{align*} n\geq n_{\varepsilon} \Rightarrow |x_{n} - x_{0}| \leq \delta_{\varepsilon} \end{align*} Gathering both results, we deduce that for every $\varepsilon > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ such that \begin{align*} n\geq n_{\varepsilon} \Rightarrow |x_{n} - x_{0}|\leq\delta_{\varepsilon} \Rightarrow |f(x_{n}) - f(x_{0})|\leq\varepsilon \end{align*} whence we get that $f(x_{n})\to f(x_{0})$, and we are done.

Let us prove the implication $(\Leftarrow)$ this time.

Let us assume that $x_{n}\in X$ converges to $x_{0}$ and $f$ is not continuous at $x_{0}$.

This means there exists an $\varepsilon > 0$ such that for every $\delta$ there corresponds a $x_{\delta}\in X$ satisfying

$$(|x_{\delta} - x_{0}| \leq \delta)\wedge(|f(x_{\delta}) - f(x_{0})| > \varepsilon).$$

In particular, if we take $\delta = 1/n$, there corresponds a $x_{n}\in X$ such that $|x_{n} - x_{0}| \leq 1/n$.

Taking the limit in the last expression, we conclude through the sandwich theorem that $x_{n}\to x_{0}$.

But $f(x_{n})\not\to f(x_{0})$ since $|f(x_{n}) - f(x_{0})| > \varepsilon > 0$, and we are done.

Hopefully this helps!

  • $\begingroup$ Thanks for time. good answer.+ $\endgroup$ Nov 28, 2020 at 20:21

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