# Precise definition of oscillation behavior of functions like $\sin(\frac1{x})$

I tried today precisely defining the oscillation behavior present in functions like $\sin(\frac1{x})$ i.e:

To do this, I started with the domain of function as in limits, and let $f(x)-L$ have infinitely many roots within any window. I came up with this:

Let $A \subseteq \mathbb{R}$, and let $a$ be a cluster point of $A$, a function $f: A \mapsto \mathbb{R}$ is said to oscillate between $b$ and $c$ ($b\neq c$) around $a$, if for any $\delta>0$ and $L\in (b,c)$ and such that: $$\text{ if } x\in A \text{ and } 0<|x-a|<\delta \text{ then } |\{x:f(x)=L\}|>M$$

Where $M$ can be any natural number. We can see that proving this property implies that the limit of $f(x)$ doesn't exist at $a$, [infact even doing so for any particular value of $L$]. Taking any $\varepsilon<\frac{|b-c|}2$ in the definition of limit, there will be no $\delta$ to satisfy the conditions. Actually, this was the motivation of doing this, I wanted at start to prove that $\lim_{x\to 0} \sin(\frac1{x})$ doesn't exists.

My questions are:

1. Is my definition alright, i.e. it explains without ambiguity what I want to define? Are there any kind of generalizations to this?
2. I am having trouble proving this for $\sin(\frac1{x})$ at $0$ between $1$ and $-1$, I know there has to be some generating function, which I visualize intuitively, but can't ultimately prove anything. Same goes for $\lceil\frac1{x}\rceil - \frac1{x}$.
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I want to make another condition that $b\le f(x) \le c$ for some open interval around $a$, so that the between word seems more correct, but I m not sure. – Sawarnik May 1 '14 at 17:45
You might be interested in en.wikipedia.org/wiki/Oscillation_%28mathematics%29 – tampis Feb 15 at 18:46

Here's a slightly simpler, equivalent definition:

$f$ oscillates between $\boldsymbol{b}$ and $\boldsymbol{c}$ around $\boldsymbol{a}$ if $$\forall \; \delta > 0 \;\; \forall \; L \in (b,c) \;\; \exists x \in A \;: \;\; 0 < |x - a| < \delta, \;\; f(x) = L.$$ (For all $\delta > 0$ and for all $L \in (b,c)$, there exists an $x \in A$ with $0 < |x - a| < \delta$, such that $f(x) = L$.)

In particular, you don't need to introduce $M$, as the specification $0 < |x - a| < \delta$ guarantees there will be infinitely many.

This makes your examples easier to prove.

• Consider $f(x) = \sin \tfrac1x, a = 0, b = -1, c = 1$. For any $\delta > 0$ and $L \in (-1, 1)$, let $\theta$ be such that $\sin \theta = L$, and pick $n$ such that $2\pi n + \theta > \tfrac{1}{\delta}$. Then $x = \frac{1}{2 \pi n + \theta}$ satisfies the required condition.

• Consider $f(x) = \lceil \tfrac1x \rceil - \tfrac1x, a = 0, b = 0, c = 1$. For any $\delta > 0$ and $L \in (0,1)$, let $n$ be such that $n - L > \frac{1}{\delta}$. Then $x = \frac{1}{n - L}$ satisfies the required condition.

You might consider replacing $(b,c)$ with an arbitrary set, and then you could say the function oscillates in that set if the given condition is satisfied.

Wikipedia defines a related notion, which assigns an "oscillation" (nonnegative real number) to a function at a point. In this case the oscillation would be $c-b$.

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Can you just replace the symbols for and, or ... with their words? Its just that I am not familiar with many of them :) – Sawarnik May 1 '14 at 18:16
Sure @Sawarnik. $\forall$ is "for all", and $\exists$ is "there exists". – 6005 May 1 '14 at 18:17