Epsilon delta proof sketch of $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$ I need to prove that $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$.
My first step was re-writing this as: $\frac{1}{1+\sqrt{x}}$. So now for the sketch of the proof i got:
Let $\epsilon>0$. Note that for all $x\in\mathbb{R}^+$ with $|x-1|<\delta$:
$\left|\frac{1}{1+\sqrt{x}}-\frac{1}{2}\right|=\frac{|1-\sqrt{x}|}{2|1+\sqrt{x}|}=\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}$. (since $|1+\sqrt{x}|\geq 0$ we can omit the absolute value symbols).
Now, observe that: $|x-1|<\delta \Leftrightarrow -\delta<x-1<\delta\Leftrightarrow -\delta+1<x<\delta+1\Leftrightarrow\sqrt{-\delta+1}-1<\sqrt{x}-1<\sqrt{\delta+1}-1$. And thus $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. (Note that has to be $\delta<1$ or the root would be negative)
Now observe that:
$\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}\leq \frac{1}{2}|1-\sqrt{x}|<\frac{1}{2}(\sqrt{\delta+1}-1)<\sqrt{\delta+1}-1<\epsilon.$
So, we get that $\delta+1<(\epsilon+1)^2\Leftrightarrow\delta<\epsilon^2+2\epsilon$.
For me,this is a bit confusing. Normally i would get a more elegant delta definition, so i don't know if what i did was true. Another problem for me was that normally i could use the $|x-1|<\delta$ directly, and maybe got another restriction. But this time i only used that $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. Can somebody clarify, if what i did was right / a way to do this more elegantly?

As $x=1$ is not considered, you are allowed to simplify the expression as
$$\frac1{1+\sqrt x}$$ when clearly tends to $\dfrac12$.
Now solving
$$\frac1{1+\sqrt x}-\frac12=\pm \epsilon$$ (the function is monotonic decreasing) we have
$$x=\left(1-\frac2{1\pm2\epsilon}\right)^2$$
and the proof works with
$$\delta\le\min\left(1-\left(1-\frac2{1+2\epsilon}\right)^2, \left(1-\frac2{1-2\epsilon}\right)^2-1\right)$$ which is the tightest bound.
