Continuity & boundedness on open interval implies uniform continuity Suppose $f(x)$ is continuous and bounded on $(0,1)$. Is $f(x)$ uniformly continuous on $(0,1)$?
I think yes, because it's bounded, i.e. there exists $M: |f(x)| < M$. We could use this M as $\delta$ in the definition of uniformly continuous function for any $\epsilon$. My textbook says, the answer is no. Why?

As David Mitra's comment suggests, $f(x) = \sin(1/x)$ over the interval $(0,1)$ is an excellent example of a function that is bounded, continuous, but not uniformly continuous. To see why, note that
$$
a_n =  \frac{1}{\pi n+ \pi/2}
$$
satisfies $a_n \to 0$ but $f(a_n) = (-1)^n$.
Perhaps this will help your intuition: if $f$ is differentiable on an open interval, then it will be uniformly continuous so long as its derivative, $f'$, is bounded.
