TITLE: $\sin(1/x)$ not uniformly continuous
QUESTION [6 upvotes]: I looked at your answer to the question posted How to prove $\sin(1/x)$ is not uniformly continuous
thank you for your helpful explanation on how to think about it. But I am failing to see why we will choose $x=\frac{1}{2πk−π/2}$ . I know that we want something $\sin(1/x)=1$ and $\sin(1/y)=−1$, but how did you get that answer. Would you mind elucidating on your motivation?
Also, I am struggling to get an intuitive idea to work with continuous uniform problems. I understand the definition, but am failing to work with it.
Thanks

REPLY [4 votes]: I think you are referring to Jonas Meyer's answer which says: 
"For what $x$ and $y$ is it true that $\sin\left(\frac{1}{x}\right)=−1$ and $\sin\left(\frac{1}{y}\right)=1$?"
Already this may be a problem as he was working with $x$ such that $\sin\left(\frac{1}{x}\right) = - 1$ but in your question, you have written $\sin\left(\frac{1}{x}\right) = 1$ (and vice versa for $y$). Is this why you don't get the stated values?
In Jonas Meyer's comment he says that $x = \displaystyle\frac{1}{2\pi k - \frac{\pi}{2}}$ and $y = \displaystyle\frac{1}{2\pi k + \frac{\pi}{2}}$. I have included the working for the $x$ values below if you are still unable to obtain them.

 $$\displaystyle\sin\left(\frac{1}{x}\right) = -1 \Leftrightarrow \frac{1}{x} = 2\pi k - \frac{\pi}{2},\ k \in \mathbb{Z} \Leftrightarrow x = \frac{1}{2\pi k - \frac{\pi}{2}},\ k \in \mathbb{Z}.$$

You should be able to find the $y$ values in a similar way.