I want to show that a specific trigonometric function is not uniform continuous, as far away from 0, it oscillates like crazy.
What I (think I) want to show: I can find an $\epsilon$ such that regardless of what $\delta$ is, I can make $|f(x) - f(y)| \ge \epsilon$ even though $|x-y| < \delta$.
What I (plan to) do: First pick an appropiate epsilon, then create sequences $x_n$ and $y_n$ so $|f(x_n) - f(y_n)| \ge \epsilon$ for all $n \in \mathbb{N} $.
Then, work on $|x_n - y_n|$, and show that it is less than some expression involvning $n$. For simplicity, say $1/n$. Then say $1/n = \delta \Rightarrow n = 1/\delta$. For such $n$, $|f(x_n) - f(y_n)|$ is still greater than or equal to $\epsilon$ (as it was for all $n$), but now $|x-y| < \delta$ for any $\delta$. And we are done.
Does above make sense? Have just started working with uniform continuity, so I am not sure if I made a mistake or two.