Prove that $f(x)=\frac{1}{x}$ is not uniformly continuous on (0,1) So I'm having difficulties understand and utilizing the definition of uniform continuity: 
$\forall \epsilon \gt 0,$ $\exists \delta>0 $ such that 
$$ |x_1-x_2|\lt \delta  \Rightarrow |f(x_1)-f(x_2)|\lt \epsilon $$
Asides from plugging in the function I'm lost as to what to do. How do I decide on which $\delta$ or $\epsilon$ to pick? (not just for this question, but for similar questions as well)
I've looked at other posts but they seem to be using techniques way too advanced for my understanding.
 A: Consider the elements $\{1/n^2 : n \geq 1\}$ of $(0,1)$. As $n$ goes to $\infty$ these elements converge to $0$ and get as close as you want. Therefore you'll satisfy the $|1/n^2-1/(n+1)^2|\leq \delta$ part. On the other hand if you take $f(1/(n+1)^2)-f(1/n^2) = 2n+1 \to \infty$. Thus, for any $\delta$ you find a counter example.

Another proof can be obtained using the answer to the following question: Prove if a function is uniformly continuous on open interval, it is continuous on closed. 
Suppose that the function $f(x) = 1/x$ is uniformly continuous on $(0,1)$. Then by the answer in the linked question, it should be extendable by continuity on $[0,1]$. This is obviously false, since the limit in $0$ is $\infty$.

Another argument: Note that uniformly continuous function maps Cauchy sequences to Cauchy sequences (this fact was used to prove the claim in the question linked above). Thus the sequence $(1/n)$ needs to be mapped to a Cauchy sequence. Since $f(1/n) = n$ is not a Cauchy sequence we can see that $f$ is not uniformly continuous. 
