# How can this theorem regarding limits of infinite sequences be proven.

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?

• 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 $c$”. Nov 28, 2020 at 20:08

Proposition

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 $$x_{0}$$.

Proof

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!

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