Prove $x^{2 \over 3} \ln(x)$ is uniformly continious Prove $x^{2 \over 3} \ln(x)$ is uniformly continuous in $(1,\infty)$
To my understanding I need to show the derivative is bounded. That will prove uniform continuity. The derivative is:  
$$ {2\ln(x) + 3} \over {3x^{1 \over 3}}$$
Now, I need to test the limit of the derivative at $\infty$ but I'm not sure how to do it. 
 A: Take $\lim_{x \rightarrow \infty} f'(x)$. It is zero as $\lim_{x \rightarrow \infty}\frac{\log(x)}{x^\alpha} = 0$ for any  $\alpha > 0$. Proof of this formula is a little lengthy. We have to use definition of $\log(x)$ obtained in the theory of Riemann integratioin. So you shall get $|f'(x)| < \epsilon$ whenever $x > G$ (depending on $\epsilon$).
So we are getting $|f'(z)| < \epsilon$ when $x > G$
Now consider $(1,G]$. See $\lim_{x \rightarrow 1} f'(x) = 1$ according to your calculation. So $f'(x)$ is bounded in $[1,G]$.
Thus ultimately we are getting $|f'(x)| < M$ when $x \in (1, \infty)$
Hope the proof is done. 
A: Let's say LHR is forbidden and as GinKin mentioned - Riemann integration is not covered in calc 1. The easiest way of proving the statement is to show that the derivative is bounded.
As AndrePoole found the derivative is $$f'(x)=\frac{2ln(x)+3 }{3x^{\frac{1}{3}}}$$
Obviously, it is bounded from below by 0 (actually, by 1, but we only need to find a boundary and 0 is easier). 
Now, let's define $x:=m^3$ (it is possible because $m\in(1,\infty)$, same as x). We can write $$f'(x)=\frac{2ln(x)+3 }{3x^{\frac{1}{3}}}=\frac{2ln(m^3)+3}{3\cdot{(m^3)^{\frac{1}{3}}}}=\frac{6ln(m)+3}{3m}=\frac{2ln(m)}{m}+\frac{1}{m}$$
Now we may use that $\forall x>0 \ : \ ln(x) \le x \rightarrow \frac{ln(x)}{x} \le 1$, thus $\frac{2 ln(m)}{m}+\frac{1}{m} < 2+1=3$.
We found that $0<f'(x)<3$, so the derivative is bounded, therefore the function is uniformly continuous for $x\in(1,\infty)$.
Good day! 
A: Use LHR on f(x). 
Derive twice to get to an expression that go to 0 as x tend to infinity (I got $\frac{2}{x^{\frac 13}}$). 
Then if $f'$ is bounded, $f$ is lipschitz thus uniformly continuous. 
