# Why $\sin\left( \frac 1 x \right)$ oscillates infinitely many times as $x \to 0$

Below, I have tried to prove why $\sin\left( \frac 1 x \right)$ starts oscillating infinitely many times as $x$ approaches zero.

We know that the Sine function is an oscillating function.

Let us assume a period $p$ such that, $$\sin\left(\frac 1 x \right) = \sin\left(\frac 1 {x+p}\right)$$

We can write, $$\frac 1 {x+p}=\frac 1 x - K$$

Where $K$ is some positive constant, $p$ is positive and $x>0$ (for simplicity[ period can be computed for both positive and negative values ]).

On rearranging we get, $$p= \frac x {1-Kx} - x$$

Here, $p$ is a function of $x$ and is therefore not constant.

Now, $$\lim_{x\rightarrow 0} p(x) = 0$$

The period falls to zero as $x$ tends to zero. The duration of each cycle becomes infinitesimal. In summary, as $x$ gets infinitely close to zero, the function oscillates infinitely many times.

I find this understanding simple. But is this approach acceptable?

• I'd suppose it would depend on your definition of "oscillating rapidly" Commented Jun 28, 2017 at 3:49
• Could you throw some more light on that?
– R004
Commented Jun 28, 2017 at 3:51
• @R004 If you want to prove something, you must first clearly articulate what it is you are trying to prove. "Oscillates rapidly" is not a mathematical statement. Much of mathematics is finding good definitions. So to try to formally spell out what you mean by "oscillates rapidly". For example, does the constantly $0$ function oscillate rapidly? Commented Jun 28, 2017 at 4:03
• I shall change the wording to avoid ambiguity. I appreciate this.
– R004
Commented Jun 28, 2017 at 4:04
• @R004 Define "oscillates". Your tweak still doesn't answer the question I posed about whether the constantly $0$ function "oscillates". Commented Jun 28, 2017 at 5:01

I believe your approach is correct.

As a little bonus, here is my explanation, that is more intuitive. Because $\frac{1}{x}$ grows from 1 to infinity as x goes from 1 to 0, $\sin(\frac{1}{x})$ must oscillate as many times within the 0 to 1 range as a regular $\sin(x)$ function must oscillate between 1 and infinity, which is an infinite number of oscillations. $\sin(\frac{1}{x})$ must fit an infinite number of oscillations between 0 and 1, thus, it has to oscillate very rapidly.You can make this argument for any number other than 1, because there must also be an infinite number of oscillations between 0 and, say, 0.0000000001, or 0.0000000000000001, or something even smaller.

• Our argument is a number for a given value of $x$. Here, the number $\frac {1}{x+p}$ for some $x$ was expressed as the difference of two numbers $\frac{1} {x}$ for some $x$ and $K$. Since I have taken all values to be positive, let me give you an example. $\frac {1}{4+1}=\frac {1}{4} - 0.05$. I don't see anything misleading here. But you could always correct me.
– R004
Commented Jun 28, 2017 at 4:16
• Intuition is a good thing and I do rely on it. But when I can, I always do the math to make understanding concrete.
– R004
Commented Jun 28, 2017 at 4:18
• Ah, sorry. I did not realize you don't have to keep K the same. No matter what you do to K, you will still have a $p$ to fix your equation, and the limit will still remain 0. I believe your approach is correct.
– D.R.
Commented Jun 28, 2017 at 4:23
• Taking $K$ to be varying with $x$ will not matter as long as $K$ tends to a number as $x$ tends to zero. Yes. $K$ becoming an infinitesimal will also not bother us. But $K$ shouldn't become infinitely big. That will make the study hopeless, I think.
– R004
Commented Jun 28, 2017 at 4:35
• Yes, but it is continuous! $\sin(x)$ oscillates an infinite number of times when x is greater than a large value, $y$, and infinity, which is obvious, right? Thus, $\sin(\frac{1}{x})$ must oscillate an infinite number of times when $\frac{1}{x}$ is between $y$ and infinity, aka when $x$ is between 0 and $\frac{1}{y}$, a tiny number.
– D.R.
Commented Jun 28, 2017 at 5:10

Your argument is correct, but you have a gap, some minor terminological issues, and the presentation is poor.

For the terminological issue, saying: "assume a period $p$ such that $\sin(1/x) = \sin(1/(x+p))$" strongly suggests that you are saying $\sin(1/x)$ is a periodic function with period $p$ which would mean $p$ is constant. Of course, as you show, $p$ is not constant, so this terminology is misleading.

The gap in your argument is when you say $\frac{1}{x}-\frac{1}{x+p} = K$ for a positive constant $K$. This can be made to be true, but you give no argument why, and the order of your presentation makes this not necessarily, let alone obviously, true. If, at this point in the argument, someone was still under the impression that $p$ was a constant, then this would be obviously false. Even if the reader hadn't assumed $p$ was constant, there's still no given reason to believe this is true.

The problem with your presentation (beyond the terminological one) is that you first assume that there is some function $p$ with the desired property, then you assume a statement which need not be true. It's not necessarily the case that if $\sin(1/x)=\sin(1/(x+p))$ that $\frac{1}{x}-\frac{1}{x+p}$ is a positive constant. You certainly give no argument for it. It's also not true for arbitrary $K$, even without knowing what $p$ is. You are not claiming this, but you give no argument any such $K$ exists. (From the perspective of first-order logic, essentially what's happening is that you are mixing around the order of quantifiers.)

Here's a rationalized version of what I think your intent was:

First, note that $\sin$ is a periodic function with period $2\pi k$ for any integer $k$. Set $K = 2\pi k$ for some positive $k$. Now $\sin(1/x) = \sin(1/x - K)$. Find $p(x)$ such that $\frac{1}{x} - K =\frac{1}{x+p(x)}$. We can do this by solving for $p(x)$ getting $p(x) = \frac{x}{1-Kx}-x$. From here proceed as the remainder of your argument.

Per MathTrain's and my comments on the question, whether this proves that $\sin(1/x)$ "oscillates rapidly" depends on what "oscillates rapidly" means.

• This makes better sense to me. I appreciate this.
– R004
Commented Jun 28, 2017 at 5:59

You have shown that if there is a period, then it must be infinitesimally small, which implies that the function under consideration must be constant. However, you don't consider the case where there is no period, which is the case that this problem falls under. Since $\sin\dfrac{1}{x}$ is not constant, then what you proved implies that $\sin\dfrac{1}{x}$ is not periodic.

Another approach is to consider the zeroes of the function $\sin\dfrac{1}{x}$.

Note that $\sin\dfrac{1}{x}=0$ if and only if $x=\dfrac{1}{k\pi}$, for some $k\in\mathbb{N}$. The sequence $a_k = \dfrac{1}{k\pi}$ approaches $0$ from above as $k\rightarrow\infty$, so then $\forall \varepsilon > 0$ there exist infinitely many $x\in(0,\varepsilon]$ such that $\sin\dfrac{1}{x}=0$. Each place that $\sin\dfrac{1}{x}$ hits zero corresponds to one half of a cycle of sine, so there are infinitely many oscillations in $(0,\varepsilon]$.

• This is, I believe, to me the most convincing reasoning. Simple and impeccable!
– R004
Commented Jun 28, 2017 at 5:58