If $f(x)$ is continuous and bounded on $(0,1)$, is it uniformly continuous on $(0,1)$? I have a question which is:

If $f(x)$ is continuous and bounded on $(0,1)$ is it uniformly continuous on $(0,1)$?
  And how does this change if $f(x)$ is also monotonic?

I have been told the that unless it's monotonic then the it is not uniformly continuous. But I don't see why...
Can someone please help me understand this? Thanks :)
 A: Let me add a little to Arturo's answer. 
First: how do you approach to problem?
You may have already known that a continuous function on any compact space is necessarily uniformly continuous. You may also know that a closed interval is a compact space. (Or you may just know that a continuous function on a closed interval is necessarily uniformly continuous.) 
Therefore, if your continuous function $f:(0,1)\to\mathbb{R}$ admits limits
$$ f_0 = \lim_{x\to 0} f(x) \qquad f_1 = \lim_{x\to 1} f(x) $$
then defining the function 
$$ \tilde{f} :[0,1] \to\mathbb{R} $$ 
by $\tilde{f}(0) = f_0$, $\tilde{f}(1) = f_1$ and $\tilde{f}(x) = f(x)$ otherwise you get a continuous function on $[0,1]$, and so $\tilde{f}$ must be uniformly continuous. And so $f$, being the restriction of $\tilde{f}$, must also be uniformly continuous. 
This tells us that if you want to look for possible problems, you must consider the case where one of the limits, say $\lim_{x\to 0}f(x)$, fail to exist.
How can the limit fail to exist? There are basically two ways: either the $\limsup$ or the $\liminf$ do not exist (in which case the function is unbounded, and which you ruled out by assumption), or the $\limsup$ and $\liminf$ both exist but do not equal one another. Using the definition of the $\limsup$ and $\liminf$ you then "immediately see" that the oscillation that Arturo mentioned can cause problems to uniform continuity. 
Note that this also explains the remark someone else made to you about monotonic functions: if you assume that $f$ is monotonic near $0$ and $1$, then you can use the fact that for bounded monotonic functions, the $\lim$ must exist, to conclude that $f$ extends to a continuous function on $[0,1]$ (like above) and therefore must be uniformly continuous. 
A: The function
$$
f(x) = \frac{2x-1}{x(1-x)}
$$
is continuous and strictly increasing on $(0,1)$ and is not uniformly continuous on that interval.
I'll tell you a non-standard analysis approach to that and let you think about the details of an $\varepsilon$-$\delta$ approach.
If $x$ is a real number between $0$ and $1$ and $dx$ is infinitely small, then $dy=f(x+dx)-f(x)$ is also infinitely small; hence $f$ is continuous on $(0,1)$.  But there are some nonstandard reals between $0$ and $1$---namely those that are infinitely close to the endpoints---where an infinitesimal change in $x$ results in a non-infinitesimal change in $f(x)$; hence $f$ is not uniformly continuous on $(0,1)$.
For an $\varepsilon$-$\delta$ approach, think of this: If you think $\delta$ is small enough, think about whether it's still small enough near the endpoints.
A: The problem is that the function may oscillate very fast between peaks and nadirs, which would spoil uniform continuity.
For an example of a continuous bounded function on $(0,1)$ which is not uniformly continuous, consider
$$f(x) = \sin\left(\frac{1}{x}\right).$$
It is certainly bounded. However, it is not uniformly continuous. Given $\epsilon=\frac{1}{4}$, for any $\delta\gt 0$ we can find a large enough value of $n$ so that 
$$\frac{2}{(2n+1)\pi} - \frac{1}{2n\pi} = \frac{4n - (2n+1)}{2n(2n+1)\pi} = \frac{2n-1}{2n(2n+1)\pi}\lt \delta,$$
yet 
$$f\left(\frac{2}{(2n+1)\pi}\right) =\sin\left(\frac{(2n+1)\pi}{2}\right) = \pm 1,$$ and $$f\left(\frac{1}{2n\pi}\right) = \sin(2n\pi) = 0,$$ so letting $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{1}{2n\pi}$, we have
$$|x-y|\lt\delta\text{ but }|f(x)-f(y)| \geq \epsilon.$$
Added. In the case of a continuous, bounded, and monotone function, however, this kind of phenomenon cannot occur. Since the function is bounded and monotone, it has a (one sided) limit at both $0$ and $1$. Extending $f$ to all of $[0,1]$ using those limits, we get a continuous function on $[0,1]$, which is therefore uniformly continuous. The restriction to $(0,1)$ gives the original function, which is there3fore also uniformly continuous. Note how this argument fails for $f(x)$ above: there is no limit as $x\to 0$. 
In fact, as noted by yoyo, it suffices (and is necessary) that the (one-sided) limits exist at $0$ and $1$ for the function to be uniformly continuous.
