Epsilon-delta proof of $\lim_{x\to\frac13^+}\sqrt{\frac{3x-1}2}=0$ 
Use the formal defintion to prove the given limit:
  $$\lim_{x\to\frac13^+}\sqrt{\frac{3x-1}2}=0$$

Not sure how to deal with $\sqrt\cdot$. Appreciate a hint.

You want to show that for any $\epsilon > 0$, there is $\delta >0$ such that if
$$ 0< x - \frac13 < \delta, $$
then 
$$  \sqrt{\frac{3x-1}{2}}  < \epsilon. $$
This is equivalent to 
$$0 < 3x - 1 < 2 \epsilon^2,$$
or 
$$ 0 <x - \frac13 < \frac23 \epsilon^2,$$
so you can choose $\delta$ based on $\epsilon$. 
