Is this correct limit proof for $\sqrt{x}$ Prove:
$\displaystyle \lim_{x\to 1} \sqrt{x} = 1$
$\displaystyle |\sqrt{x} - 1| < \epsilon \space \text{such that}\space |x - 1| < \delta$ 
Let $|x - 1| < 1 \implies |x| < 2 \implies -2 < x < 2$
$\sqrt{x} - 1 \implies \sqrt{x} - 1 < \sqrt{2} - 1$
$|\sqrt{x} - 1| < \sqrt{2} -1 $
$-2 < x  < 2$
$ -3 < x - 1 < 1 \implies 1 < |x - 1| < 3$ 
$1 > \sqrt{2} - 1$
$\therefore |\sqrt{x} - 1| < \sqrt{2} - 1 < 1 < |x - 1| < \delta$
Finally, $|\sqrt{x} - 1| < |x - 1| < \delta$
Therefore, $\delta = \min(1, \epsilon)$  $\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \blacksquare $
 A: The elements of a proof are there, but the logic is not clear. 
We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-1|\lt \delta$ then $|\sqrt{x}-1|\lt \epsilon$.
Let $\epsilon\gt 0$ be given. Let $\delta=\min(1,\epsilon)$. If $|x-1|\lt \delta$, then $x\gt 0$, and
$$|\sqrt{x}-1|=\left|\frac{x-1}{\sqrt{x}+1}\right|\lt |x-1|\lt \epsilon.$$
A: As André said, make your proof a bit clearer and it will work.
Since you haven't expressed the necessity of introducing $\delta$, you might want to consider the following simpler alternative: $$\displaystyle \left\lvert \sqrt{x}-1 \right\rvert < \epsilon \\-\epsilon<\sqrt{x}-1<\epsilon \\ (1-\epsilon)^2<x<(1+\epsilon)^2 \\ 1-2\epsilon+\epsilon^2<x<1+2\epsilon+\epsilon^2.$$
A: In this case you can directly solve the inequation
$$
|\sqrt{x}-1|<\varepsilon
$$
where $\varepsilon>0$. The inequation is equivalent to
$$
\begin{cases}
\sqrt{x}<1+\varepsilon\\
\sqrt{x}>1-\varepsilon
\end{cases}
$$
If $\varepsilon>1$, the solution set contains the interval $[0,(1+\varepsilon)^2)$ which is a neighborhood of $1$.
If $0<\varepsilon\le1$, the system of inequations is equivalent to
$$
(1-\varepsilon)^2<x<(1+\varepsilon)^2
$$
which is again a neighborhood of $1$.
If you absolutely need a $\delta>0$, you can take $\delta=1$ for $\varepsilon>1$ and
$$
\delta=\min\{(1+\varepsilon)^2-1,1-(1-\varepsilon)^2\}
$$
for $0<\varepsilon\le1$. Since $(1+\varepsilon)^2-1=2\varepsilon+\varepsilon^2$ and $1-(1-\varepsilon)^2=2\varepsilon-\varepsilon^2$, you see that $\delta=2\varepsilon-\varepsilon^2$ is a good choice.
