Proving continuity of $f(x)=x\cos(2\pi/x)$ at $x=0$

Question

I know that the function $f(x)=x\cos(2\pi/x)$ if $x\neq0$ and $f(0)=0$ is continuous at $x=0$ using $\epsilon-\delta$ as follows:

$\lvert x\cos(2\pi/x)\rvert=\lvert x\rvert\lvert\cos(2\pi/x)\rvert\leq\lvert x\rvert<\delta$, so taking $0<\delta<\epsilon$ it is already proven.

However, there should be another way, using that $f$ continuous at $x_0$ by definition if $\forall\;V\in \mathcal{N}_{f(x_0)}\;,\;f^{-1}(V)\in\mathcal{N}_{x_0}$.

I consider the local basis $\beta_{f(0)}=\beta_{0}=\{(-r,r)\mid r>0\}$. Let $V=(-r,r)$, $f^{-1}(V)=\{x\in\mathbb{R}\mid -r<x\cos(2\pi/x)<r\}$. How would I compute $f^{-1}(V)$ so I can deduce that $f^{-1}(V)\in\mathcal{N}_0$, that is, how would I isolate $x$?

Answer 1

You have already proved that $$ f^{-1}(V)\supseteq V $$ with your $\varepsilon$-$\delta$ proof. So $f^{-1}(V)$ is a neighborhood of $0$.

You don't need to know precisely what's $f^{-1}(V)$, but just that it includes a set of the form $(-s,s)$, for some $s>0$.

By the way, since $$ -|x|\le x\cos\frac{2\pi}{x}\le |x| $$ for all $x\ne0$, the squeeze theorem says that $$ \lim_{x\to0}x\cos\frac{2\pi}{x}=0 $$ which is all you need.