Proving $\lim_{ x\to 0 } \sin \frac{2\pi }{x}$ does not exist

Question

Here's my attempt:

Suppose there is some $L \in \mathbb{R}$ such that $\lim_{ x\to 0 } \sin \frac{2\pi }{x} = L$. Then, if we let $\varepsilon = 1$, there would exist $\delta > 0$ such that $\left| f(x) - L\right| < 1$ for all $0< |x| < \delta $. Now, by the Archimedean property, there is $n \in \mathbb{N}$ such that $\frac{1}{n} < \delta $. Pick $x_1 = \frac{4}{4n+1}$ and $x_2 = \frac{4}{4n+3}$ and note that $x_1 , x_2 < \delta $. Now for $x_1$, we have

$ \left| f(x_1) - L \right| = \left| \sin \left( (4n+1) \frac{\pi}{2} -L \right) \right| = \left| 1-L \right| < 1 $

And similarly

$ \left| f(x_2) - L \right| = \left| \sin \left( (4n+3) \frac{\pi}{2} -L \right) \right| = \left| -1-L \right| < 1 $

Thus, L is within one unit of both $-1$ and $1$. However, there cannot be any such number. A contradiction!

Is this proof okay?

Answer 1

The only mistake I can find is that $x_1 , x_2 < \delta$ ought to be $0<x_1 , x_2 < \delta$. Also, depending on who's grading, "there cannot be such a number" might need some proof or justification. Or it might not. Apart from that, well done!

Answer 2

As an alternative simply note that for $n\to \infty$

• $x_n=\frac4n \to 0 \implies \sin \frac{2\pi}{\frac4n}=\sin \frac{\pi n}2=\pm 1$

therefore the given limit doesn’t exist.