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.
 A: 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 $\epsilon$. 
A: $\forall \epsilon>0,\exists \delta>0\text{ such that } 0<(x-\frac 13)<\delta \implies |\sqrt{\frac{3x-1}{2}}|<\epsilon$
$\sqrt{\frac{3x-1}{2}} < \sqrt{\frac 32} \sqrt \delta$
$\delta \le \frac 23 \epsilon^2\implies|\sqrt{\frac{3x-1}{2}}|<\epsilon$
A: By the axioms of order of a field we know that multiplying both sides of an inequality by a positive quantity the inequality doesnt change, then
$$\left|\sqrt{\frac{3x-1}2}\right|<\epsilon\iff 
\left|\frac{3x-1}2\right|<\epsilon\cdot \left|\sqrt{\frac{3x-1}2}\right|<\epsilon^2$$
where the last inequality comes from the application again of the LHS. Then we finally have that
$$\left|\frac{3x-1}2\right|<\epsilon^2\iff |3x-1|<2\epsilon^2$$
From here it is easy to apply the definition of limit and see that $\lim_{x\to 1/3^+} (3x-1)=0$ (the expression $2\epsilon^2$ only depends of epsilon, what can be chosen arbitrarily).
And then because the function $f(x)=\sqrt x$ is continuous at zero we have that
$$\lim_{x\to 1/3^+}\sqrt{\frac{3x-1}2}=\sqrt{\frac {\lim_{x\to 1/3^+} (3x-1)}2}=\sqrt{\frac02}=0$$
