Proving a function is not uniformly continuous I have a homework question to which is 

If $f(x)$ is diferentiable in $(0,\infty)$  and $f'(x)>\frac{1}{x}$
  for every $x>0$ then $f$ is not uniformly continuous in $(0,\infty)$.

The question has a clue which asks me to prove that for $0<a<b$ then $f(b)-f(a)>\frac{b-a}{b}$ which I have also not managed to prove.
But even if I did I can't how to use that fact to solve the question. 
Can someone help me out? Thanks a lot :) 
 A: We'll first prove the hint. 
Let $0 <a<b$.
For this, we use the mean value theorem in $[a,b]$.
Why does $f$ satisfy the hypothesis of MVT on $[a,b]$?
As $f$ is differentiable on $(0,\infty)$, $f$ is continuous on $[a,b]$. And, $f$ is, in fact, differentiable on $[a,b]$, if you define differentability at endpoints using one-sided derivatives.
So, there exists $c \in (a,b)$ $$\dfrac{f(b)-f(a)}{b-a}=f'(c) >\dfrac{1}{c}>\dfrac{1}{b}$$ Now, noting that, $b-a>0$, we can actually take $b-a$ to the other side, taking us to $$f(b)-f(a)>\dfrac{b-a}{b}$$ 
This proves your hint.
Now to prove that, $f$ is not uniformly continuous, we actually prove the contrapositive of the definition:
Contrapositive of uniform continuity

$f$ is NOT uniformly continuous if there exists  $\epsilon>0$ such that for each $\delta >0$ there exists $x,y$ such that $|x-y|<\delta$ but $|f(x)-f(y)|\geq \epsilon$.

I think this should not be hard. So, how do we go about this?
Set $b=2a$. Note that this is consistent with our assumption on $a$ and $b$ (i.e. $a < b $). There exists an $\epsilon>0$, here $\epsilon =\dfrac{1}{2}$, such that for any $\delta>0$, here any $a>0$, the claimed inequality holds.
So, we are through.
A: Hints: 
1) Use the Mean Value Theorem to obtain the inequality.
2) Set $b=2a$. In this case $f(b)-f(a)>1/2$. So, can you choose $a$ and $b$ as closely as you like and still have $|f(b)-f(a)|$ big?
