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I have given a periodic function ($f(x+1)=f(x)$) $$f(x)=\begin{cases}4x&(x\in[-0.25;0.25))\\2-4x& (x\in[0.25;0.75])\end{cases}$$

and the function $$g:\mathbb R\rightarrow\mathbb R, x\mapsto\begin{cases}f\left(\frac1x\right)&(x\neq0)\\0&(x=0)\end{cases}$$

So I want to prove $g$ oscillates in $0$ but I am really stuck into it. How can you prove that $g$ is taking on every value of $[-1;1]$ in any $\varepsilon$-neighborhood of $0$ ? And why is $g$ bounded by $\pm1$ which you need for $[-1;1]$?

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up vote 4 down vote accepted

In every $\varepsilon$-neighbourhood of $0$, there is an interval $\left[\frac1{n+1},\frac1n\right]$, hence all values $f(x)$ with $x\in[n,n+1]$ occur. And $g$ is bounded because $f$ is.

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thanks a lot! but I still don't get why $f(1/x)$ is bounded by -1 and 1? – user58679 Jan 18 '13 at 12:42
$f(\frac1x)=f(y)$ with $y=\frac1x$ and $f(y)\in[-1,1]$ for all $y\in\mathbb R$ - at least this is true for $0\le y\le1$ by inspecting and for all other $y$ by periodicity. – Hagen von Eitzen Jan 18 '13 at 12:48

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