proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$. I'm having trouble with proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$.
So far I have:
$\frac{1-\sqrt{x}}{1-x} =\frac{1}{1+\sqrt{x}} $
If $x\in dom (\frac{1}{1+\sqrt{x}}$) and $|x-1| <\delta$ then $ |\frac{1}{1+\sqrt{x}}- \frac{1}{2} |< \epsilon$
So I started writing out $|\frac{1}{1+\sqrt{x}}- \frac{1}{2}|$, and got that it's equal to$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|$. And since we have $|x-1| <\delta$ ,
$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|< |\frac{\delta}{2(1+2\sqrt{x}+x)}|$. 
I'm kind of stuck here. Any help on how to continue is much appreciated!

So you arrived at:
$$\left|\frac{1-x}{2(1+2\sqrt{x}+x)}\right|
=\frac{\left|1-x\right|}{2\left|1+2\sqrt{x}+x\right|}$$
and $\delta$ gives you an upper bound for the numerator.
Hint: note that $x>0$ (domain) and look for an (obvious) lower bound for the denominator.
