proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$.

Question

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!

Answer 1

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.

Answer 2

Hint: for $x>0$:

$|\frac{1-x}{2(1+2\sqrt(x)+x)}| \le |\frac{1-x}{2}|$