I have to prove that $\frac1x $ with $x \in (0,\infty)$ is not uniformly continuous. I understand that the further you approach the $0$ the greater your $e$ and the smaller your $\delta$ becomes (in the limit even $\delta = 0 $ ?).
But I need help with the "technical" proof, can someone give me an instruction to do this?
On
The definition of a uniformly continuous function $f$ is $$\forall \varepsilon > 0 ~ \exists \delta > 0 ~ \forall x,y \in D(f) : |x-y|<\delta \Rightarrow |f(x) - f(y)| < \varepsilon.$$ (For all $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x)-f(y)|<\varepsilon$ for all $x,y$ that satisfy $|x-y|<\delta$)
To show that $f$ is not uniformly continuous we can show that the logical negation of this definition holds. This would be
$$\exists \varepsilon > 0 ~ \forall \delta > 0 ~ \exists x,y \in D(f) : |x-y| < \delta \wedge |f(x)-f(y)| \geq \varepsilon$$
(There exists an $\varepsilon > 0$ such that for all $\delta > 0$ there are $x,y$ with $|x-y| < \delta$ but also $|f(x)-f(y)| \geq \varepsilon$)
Let $f(x) = 1/x$ and $x_n = 1/n$. Then we have $$|f(x_n) - f(x_{n+1})| = |n - (n+1)| = 1$$ and $|x_n - x_{n+1}| = 1/(n^2+n)\to 0$. This means that if we choose $\varepsilon = 1/2$ and are given any $\delta > 0$, we can find a sufficiently large $n \in \mathbb{N}$, such that $|x_n - x_{n+1}| < \delta$ and $$|f(x_n) - f(x_{n+1})| = 1 > \frac{1}{2} = \varepsilon$$ hold. It follows that $f$ is not uniformly continuous.
The main important facts about $f(x) = 1/x$ is that it gets arbitrarily large as $x \to 0$ and it's monotonically decreasing. Those mean that for any number $N$, you can find $t > 0$ such that $f(x) > N$ for every $x \in (0, t]$.
Now suppose that $f$ were uniformly continuous, that is, for any $\varepsilon > 0$ there is $\delta > 0$ such that for $|x - y| < \delta$, we get $|f(x) - f(y)| < \varepsilon$. Choose $\varepsilon = 1$; really, any positive number will do. Here let $y = \delta$ and define $Y = f(y) = f(\delta)$. Then we can find $t > 0$ such that $f(x) > Y + 1$ for any $x \in (0, t]$. Clearly then $t < \delta$ or else the statement would be violated by choosing $x = \delta$. Now choose an arbitrary $x \in (0, t]$. Because $t < \delta$, we have $$ |x - y| < y = \delta $$ while $$ |f(x) - f(y)| = f(x) - f(y) > Y + 1 - Y = 1 = \varepsilon $$ (using that $f$ is monotone). So actually it's not uniformly continuous.
The moral of the story is that I didn't use any properties of $f$ except that it goes off to infinity in a finite interval, and it's monotone, so no such function can be uniformly continuous. You don't even need the latter property, it's just slightly more convenient; you should try eliminating it from this proof.