Proving that $f(x) = \frac{1}{x}$ is not uniformly continuous over $(0,1)$ - Approach to choosing the correct $x$ and $y$ I've looked at the various other posts on this question and I've also worked on uniform continuity in the past. But right now as I work through Spivak's Calculus, I'm seeing that I haven't actually learned how to find choices for my $x$ and $y$ elements in a structured well thought out way. In the past what I thought was doing this was actually me just regurgitating what a professor did in their proof without any understanding of how $x$ and $y$ were chosen.
To illustrate where I'm stuck:
Let $\epsilon = 1$. Taking arbitrary $\delta &gt; 0$. I encounter the following:
$$\bigg|\frac{1}{x} - \frac{1}{y} \bigg| &gt; \epsilon = 1$$
I've seen prior choices such as letting $x = \delta$ and $y = \frac{\delta}{2}$ or letting $x &lt; \delta$ and $y = \frac{x}{2}$. But I want to understand how these choices were made initially.
I know that $x(\delta)$ and $y(\delta)$ as well if I fiddle with the the expression above I have
$$\bigg|\frac{x-y}{xy}\bigg|$$
I also know that since $x,y \in (0,1)$ that $xy &lt; 1$ (but this now won't help me when I want to extend the interval to $(0, \infty)$.
It appears I'm throwing a lot of stuff at the wall hoping for something to stick and this isn't very conducive. How should I be reasoning these steps out?
Some rough steps :

Visualize the graph of this function : since you are asked to prove that the function is not uniformly continuous, you must look for a place where the function is increasing really fast (or has steep slope) . (Note : a heuristic is that this can happen near a point where the function takes the value infinity : that point won't be there in the domain, but points close to it will be there)

Why? Because, at these points, the $x$ coordinates are really close, but the $y$ coordinates are far apart. Remember uniform continuity says that the $y$ - coordinate difference between two points on the function's graph can be uniformly controlled by controlling their $x$-coordinate difference : so the points lying on these high-slope regions are most likely to provide a contradiction.

After doing this, look for points $x,y$ for which the expression for $f(x)$ is very simple. For example, if $f(x) = \sin x$, then for $x$ being multiple of $\frac {pi }4$, for example, the expression is simple.

Sometimes, you will be able to do both : find a sequence of points in the slopy region, for which the function values are really easy to evaluate. Then you can compare these values and come to a contradiction.



Let us apply these principles for $f(x) = \frac 1x$. Drawing $\frac 1x$ in your head, you see that if $x = 0$ the expression is infinite. So points near $0$ is where you can try to look.
Note that if $x$ is the reciprocal of an integer, then $f(x) $ is that integer! So taking $x = \frac 1n$ for $n \in \mathbb N$ gives simple values of $f(x)$. Thus, we don't need to complicate matters by looking at the general expression $|f(x)-f(y)|$.
Finally, we notice that if $x_n = \frac{1}{n}$ and $y_n = \frac 1{n+1}$, then $|x_n-y_n|$ can be made small enough, but we always have $|f(x_n)-f(y_n)| = 1$. So, for a given $\delta$, we can easily take such large $N$ that $\frac 1N &lt; \delta$ (and $|x_N - y_N| =x_N-y_N &lt; x_N &lt; \delta$) but $|f(x_n)-f(y_n)| = 1$.
Thus uniform continuity is contradicted with $\epsilon = 1$, because we can find for any $\delta$ a pair of points differing by less than $\delta$ but whose function values have difference $1$.
You can keep these points in mind while approaching &quot;non-uniform continuity&quot; questions.
Simply let $ x = 2y \implies \left|\dfrac{x-y}{xy} \right| = \dfrac{1}{x} \ge \epsilon = 1$ when $0 &lt; x &lt; \delta &lt; 1$.