What is a function that is both bounded and continuous on $(1,2)$ but not uniformly continuous on the interval? I have been seriously thinking about what I could use to answer this question and I just can't think of anything. I have tried doing $\frac{1}{x}$, $e^x$ and even considered $\tan(x)$. Maybe $\sin(\frac{1}{x})$ could work? I would appreciate any help!
 A: The function is bounded so there is no infinite limit possible at the lower or upper bound of the interval (i.e in $1$ or $2$).
If you can extend the function by continuity to the closed interval $[1,2]$, this will make it continuous on a compact set and therefore uniformly continuous.
So you are searching for a function that is continuous on $(1,2)$ but with no one side limit in $1^+$ or in $2^-$, since you are constrained by boundedness, it means that $f(x)$ should take infinitely many times different values whatever small the interval $[1,1+\epsilon)$ is, and since it is continuous, $f$ has to oscillate very quickly between some extremum values.
One such example is $f(x)=\sin(\frac 1{x-1})$ which has the whole segment $y\in[-1,1]$ as an adherent value for $y=f(x)$ in neighborhood of $1^+$, therefore not uniformly continuous.

Edit: answering your question in comment
The delta is not very important, just take $\epsilon=\frac 12$ for instance there is always an $x_n=1+\frac 1{(2n\pm\frac 12)\pi}$ such that $f(x_n)=\pm 1\notin[\ell-\epsilon,\ell+\epsilon]$ and $|x_n-1|<\delta$ for $n$ large enough.
i.e. you found two values more than $2\epsilon$ apart, so it is impossible to find a suitable potential limit $\ell$ such that $|f(x_n)-\ell|<\epsilon$.
