TITLE: Is any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous
QUESTION [7 upvotes]: Question is to check if  :
any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous.
I know that any continuous function on compact subset of $\mathbb{R}$ is uniformly continuous.
As $(0,1)$ is not compact, we can not say anything at this time.
Now, as it is given that $f$ is differentiable, if its derivative $f'$is bounded then  $f$ is uniformly continuous.
So, I am trying to look for differentiable functions $f$ on $(0,1)$ such that $f'$ is unbounded.
i am not very familiar with large number of differentiable functions with unbounded derivatives.
I know $f(x)=\sqrt{x}$ has unbounded derivative, but $\sqrt{x}$ is uniformly continuous....
So, I would like someone to help me out with some hint.
P.S : I have just now saw one example 
$$f(x)=x^2\sin{\frac{1}{x^2}}$$
which  is differentiable but is not bounded.
I see that $\sin(\frac{1}{x^2})$ is bounded by $1$ and if $x\in (0,1)$ then so is $x^2$ and so is $x^2\sin{\frac{1}{x^2}}$
So, $f(x)=x^2\sin{\frac{1}{x^2}}$ is from $(0,1)$ to $[0,1]$ whose derivative is unbounded.
Now, the problem reduces to show that $f(x)$ is not uniformly continuous... :(

REPLY [4 votes]: Consider the function $f:(0,1) \to [0, 1]$ be defined by $f(x)=(1+\cos(1/x))/2$ on $(0,1)$.
clearly it's differentiable and $f'(x)=-\cos(1/x)/2(x^2)$ for  $0< x <1$.
since,
Any continuous function  $f:(a,b) → \mathbb R$ is uniformly continuous on bounded open interval $(a,b)$ iff  $f(a+)$ and $f(b-)$ both exist.
Here $f(1/(2n−1) \pi)=0$ and $f(1/2n \pi)=1$ for $n=1,2,3, \ldots$.
so $f(0+)$ does not exist.
Hence $f(x)$ is not uniformly continuous on $(0,1)$ .