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.

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. 
