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 $0 < x - \sqrt{3} < \delta$?
 A: We want to show that for all $\epsilon > 0$, there exists a $\delta > 0$ such that $x- \sqrt{3} < \delta$ (this is where the $\sqrt{3}^+$ comes in!) implies $\sqrt{x^2 - 3} < \epsilon$.
Let $\epsilon > 0$. Pick $\delta = \min(1, \frac{\epsilon^2}{1+ 2\sqrt{3}})$. 
Assume that $x - \sqrt{3} < \delta$. Then we have $x+ \sqrt{3} < \delta + 2\sqrt{3}$, so taking their product,
$$ x^2 - 3 < \delta(\delta + 2\sqrt{3}) .$$
Sinc $\delta = \min(1, \frac{\epsilon^2}{1+ 2\sqrt{3}})$, in particular $\delta \leq 1$. So $\delta + 2\sqrt{3} \leq 1 + 2\sqrt{3}$, which means that
$$ x^2 - 3 < \delta (1 + 2\sqrt{3}) $$
Now use that $\delta \leq \frac{\epsilon^2}{1+ 2\sqrt{3}}$ to get
$$ x^2 - 3< \epsilon^2$$
and thus $\sqrt{x^2 - 3} < \epsilon.$ 

Edit: Here's some motivation behind this real analysis black magic.
You're given $x - \sqrt{3} < \delta$ and want $\sqrt{x^2 - 3} < \epsilon$. First of all, notice that $\sqrt{x^2 - 3} = \sqrt{x - \sqrt{3}}\sqrt{x + \sqrt{3}}$. The $\sqrt{x - \sqrt{3}}$ goes to $0$, but the $\sqrt{x + \sqrt{3}}$ term doesn't. It sort of hovers around $\sqrt[4]{3}$. Ideally, we'd like to do something like the following:
$$ x - \sqrt{3} < \delta \implies (\sqrt{x - \sqrt{3}})\sqrt{x + \sqrt{3}} < \epsilon .$$
So it seems natural to "solve" for the part we know, $x - \sqrt{3}$, to get
$$ x - \sqrt{3} < \frac{\epsilon^2}{x + \sqrt{3}} $$
Now this should start to look familiar. We're almost done! If we would be able to set
$$ \delta := \frac{\epsilon^2}{x + \sqrt{3}} $$
we'd have a proof. Unfortunately we need to get rid of the $x$ in the denominator.
Specifically, we need some lower bound for $\frac{1}{x + \sqrt{3}}$ that's a number independent of $x$. Luckily, since $x$ is "close" to $\sqrt{3}$, $\frac{1}{x + \sqrt{3}}$ won't grow without bound anywhere. To bound this denominator, we'll additionally require that $\delta \leq 1$. This gives us
$$ \frac{1}{x + \sqrt{3}} \geq \frac{1}{1 + 2\sqrt{3}},$$
so by requiring
$$ \delta = \frac{\epsilon^2}{1 + 2\sqrt{3}} $$
our proof will stay correct. 
The way to formally write this is 
$$ \delta := \min(1, \frac{\epsilon^2}{1 + 2\sqrt{3}}) .$$
A: The statement appears somewhat confused. This limit would be equivalent to showing for any $\epsilon > 0 $ there is a $\delta >0$ such that $|\sqrt{x^2-3}| < \epsilon$ whenever $0 < x-\sqrt{3}  < \delta$.
Some hints: $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$ and $|x+y| \leq |x|+|y|$.
A: $\lim_{x \to \sqrt{3}^+}\sqrt{x^{2}-3}=\lim_{x \to \sqrt{3}^+}\sqrt{(x-\sqrt3)(x+\sqrt{3})}=\lim_{x \to \sqrt{3}^+}\sqrt{x-\sqrt3}\sqrt{x+\sqrt{3}}$  
Now when $x \approx \sqrt{3}$  
$\sqrt{x+\sqrt3}\sqrt{x-\sqrt{3}}\approx\sqrt{2\sqrt3}\sqrt{x-\sqrt3}$  
Now consider what happens when $x\rightarrow\sqrt{3}^+$
A: 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.
