TITLE: Why $\sin\left( \frac 1 x \right) $ oscillates infinitely many times as $x \to 0$
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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 $(0,\varepsilon]$.