How to show $\lim_{x \to \sqrt{3}^+}\sqrt{x^{2}-3}=0$ I understand that given any $\epsilon>0$, I need to find a $\delta > 0$ such that $|f(x) - \sqrt{3}|<\epsilon$ whenever $0 < x - \sqrt{3} < \delta$.
Starting with $|\sqrt{x^{2}-3}|$, how do I simplify this to show this is $< \epsilon$ for an appropriately choosen $\delta$ using $0 < x - \sqrt{3} < \delta$?

We have that $0<x-\sqrt{3}< \delta$.
Take $\delta<1$
Thus $|\sqrt{x^2-3}|=|\sqrt{x+\sqrt{3}}||\sqrt{x-\sqrt{3}}|<|\sqrt{2 \sqrt{3}+ \delta}||\sqrt{x-\sqrt{3}}|< \sqrt{2 \sqrt{3}+ \delta}\sqrt{\delta}<(\sqrt{2 \sqrt{3}+ 1}) \sqrt{\delta}$
Finaly you can take $\delta=\min\{\frac{\epsilon^2}{\sqrt{2 \sqrt{3}+1}},1\}$ and you are done.
