Can we prove that if $\lim_{n \to \infty} a_n = c$, and $f$ is continuous, then $\lim_{n \to \infty} f(a_n) = f(c) $?

Question

Good day everyone,

I was wondering how we could say that if $\lim_{n \to \infty} a_n = c$, and f is continuous, then $\lim_{n \to \infty} f(a_n) = f(c) $

I found this in the bisection method used to prove Bolzano's theorem. Do you know if there is a theorem (and which) or a corollary that explains the above statement? Maybe the reason is obvious but I still do not get it; if anyone could help me I would be so grateful.

Thanks in advance!

Answer 1

Absolutely! There exists $\delta>0$ such that whenever $|x-c|<\delta$, we have $|f(x)-f(c)|<\epsilon$. Since $x_n\to c$, there exists $N\in\mathbb{N}$ such that whenever $n\geq N$, $|x_n-c|<\delta$. Therefore, for all $n\geq N$, $|f(x_n)-f(c)|<\epsilon$. This is what it means to say that $f(x_n)\to f(c)$.

Answer 2

you can show a if and only if, that is;

A function $f:A\subset \mathbb{R} \rightarrow \mathbb{R}$ is continuos at the $c \in A$ if and only if for every sequence $\left\{x_{n}\right\}$ in $A$ such that $\left\{x_{n}\right\} \rightarrow c$, the sequence $\left\{f(x_{n}\right\} \rightarrow f(c)$

proof.$\Leftarrow )$ Suppose that $lim_{x \to c}f(x)=f(c)$ it is false, that is:

$(\exists \epsilon >0)(\forall \delta>0)(|x-c|<\delta \wedge \ |f(x)-f(c)| \geq \epsilon)$, hence some $n \in \mathbb{N}$ there is $x_{n}$ such that $|x_{n}-c|< \frac{1}{n}$ but $|f(x_{n}-f(c)|\geq\epsilon$, which implies that $\left\{f(x_{n}\right\} \nrightarrow f(c)$, the which is a contradiction.

Hence $f$ is continuos.