The way in which I solved it doesn´t lead to the same answer as the answer book so I wanted to know if it´s a valid answer.
Problem:
Find a $\delta$ shuch that $0<|x-1|<\delta\Rightarrow|\sqrt{x}-1|<\epsilon\;\;\;\;$ $\;\forall\epsilon\in(0,1)$.
My solution:
$$|\sqrt x-1|<\sqrt x+1$$ thus $$\sqrt x+1<\epsilon\Rightarrow|\sqrt{x}-1|<\epsilon$$ $$|x|<\epsilon^2-2\epsilon+1\Rightarrow |\sqrt{x}-1|<\epsilon \;\;\;\;\;\;\;\;\;\text{(1).}$$ If $|x|<\epsilon^2-2\epsilon+1$ is true, then $|\sqrt{x}-1|<\epsilon$ is true, we just have to find a $\delta$ such that, $\forall x$, $|x|<\epsilon^2-2\epsilon+1$.
We also have $$\delta>|x-1|\geq|x|-1$$ thus $$|x|-1<\delta$$ $$|x|<\delta+1\;\;\;\;\;\;\;\;\;\;\text{(2)}$$ therefore, in order for $|x|<\epsilon^2-2\epsilon+1$ to be true$\;\forall x$, by (1) and (2), we must have a $\delta$ such that $$|x|<\delta+1<\epsilon^2-2\epsilon+1$$ so $$\delta<\epsilon^2-2\epsilon$$ but $\epsilon^2-2\epsilon<0\;\forall\epsilon\in(0,1)$ which leads to $$\delta<0$$ but $\delta>0$, a contradiction.
Where did I go wrong?
Thanks in advance.
Before (1), $\sqrt x+1<\epsilon\Rightarrow|\sqrt{x}-1|<\epsilon\;$ is stated. But $\sqrt x+1<\epsilon$ can not be true for any $x$ because the smalest value for epsilon would be at $x=0$ which gives $$1<\epsilon$$ which, of course can´t be true since $\epsilon\in(0,1).$
Thanks to @NoName for pointing this out.