What is wrong? $\epsilon$-$\delta$ proof that $\lim \limits_{x \to 1} \sqrt{x}=1$

Question

$\epsilon$-$\delta$ proof that $\lim \limits_{x \to 1} \sqrt{x}=1$

My proof:

We have that $0 < \lvert x-1 \rvert < \delta$ and $\rvert \sqrt{x}-1 \lvert \lt \epsilon$

$\lvert x \rvert -1\le\lvert x-1 \rvert \lt\delta$ which gives $\lvert x\rvert \lt \delta +1$

$\rvert\sqrt{x} \lvert -1 \le \rvert \sqrt{x}-1 \lvert \lt \epsilon$ Which gives $\rvert\sqrt{x} \lvert\lt \epsilon +1$ squaring and taking the absolute value gives us $ \rvert x \lvert \lt \lvert \epsilon^2 + 2\epsilon +1 \rvert$

Now simply let $\delta = \lvert \epsilon^2+2\epsilon +1\rvert-1$

There is something wrong in this proof,i suspect it's somewhere with me squaring the second inequality but i'm not sure.

Second correct proof:

We still have that $0 < \lvert x-1 \rvert < \delta$ and $\rvert \sqrt{x}-1 \lvert \lt \epsilon$

Also $$\lvert \sqrt{x}-1\rvert \lt \epsilon$$ $$\therefore -\epsilon \lt \sqrt{x}-1 \lt \epsilon$$ $$\therefore 1-2\epsilon +\epsilon^2 \lt x \lt 1+2\epsilon +\epsilon^2$$

Then

$$\rvert x-1 \lvert \lt \min(\lvert-2\epsilon +\epsilon^2\rvert, \lvert2\epsilon +\epsilon^2\rvert)$$

Now let $\delta = \min(\lvert-2\epsilon +\epsilon^2\rvert, \lvert2\epsilon +\epsilon^2\rvert)$

Which is correct. Given the similarity to the two solutions in the proofs there must be something small that's missing in the first proof, but I don't know what. Any help?

Answer 1

$$\sqrt x-1=\frac{x-1}{\sqrt x +1}$$Take absolute values and bound the denominator below by 1.