$\epsilon$-$\delta$ proof that $\lim_{x \to 1} \sqrt{x} = 1$ I'm trying to teach myself how to do $\epsilon$-$\delta$ proofs and would like to know if I solved this proof correctly. The answer given (Spivak, but in the solutions book) was very different.

Exercise: Prove $\lim_{x \to 1} \sqrt{x} = 1$ using $\epsilon$-$\delta$.
My Proof:
We have that $0 < |x-1| < \delta $.
Also, $|x - 1| = \bigl|(\sqrt{x}-1)(\sqrt{x}+1)\bigr| = |\sqrt{x}-1||\sqrt{x}+1| < \delta$.
$\therefore |\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|}$
Now we let $\delta = 1$. Then
\begin{array}{l}
-1<x-1<1 \\
\therefore  0 < x < 2 \\
\therefore  1 < \sqrt{x} + 1<\sqrt{2} + 1 \\
\therefore  \frac{1}{\sqrt{x} + 1}<1.
\end{array}
We had that $$|\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|} \therefore |\sqrt{x}-1|<\delta$$
By letting $\delta=\min(1, \epsilon)$, we get that $|\sqrt{x}-1|<\epsilon$ if $0 < |x-1| < \delta $.
Thus, $\lim_{x \to 1} \sqrt{x} = 1$.

Is my proof correct? Is there a better way to do it (still using $\epsilon-\delta$)?
 A: The proof is correct but can be simplified. You don't need the part "Now let $\delta=1$...". In fact it is always true that
$$
  \frac{1}{\sqrt x + 1} \le 1
$$
since $\sqrt x \ge 0$.
Also, a matter of style. In the first line you don't have $0 < |x-1|<\delta$ but you suppose it (this is because $\delta$ is not already been given, but has to be found yet). The same when you write "let $\delta = 1$" you should write "if $\delta \le 1$ ..."
A: Your proof is correct.
We can also adopt the following:
Since $|\sqrt x-1|\lt \epsilon$ is equivalent with $1-2\epsilon+\epsilon^2\lt x\lt 1+2\epsilon+\epsilon^2$, we can choose $\delta$ so that $0\lt\delta\lt \min\{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|\}$.
Hope this helps.  
A: Using your work here is another flavor of this proof:
Let $\epsilon >0$, and put 
 $\delta= \epsilon(\sqrt{x}+1)$.
Assume $0<|x−1|<\delta$.
Then
     $$|F(x)−L|=|x−1|
             =∣(\sqrt{x}−1)(\sqrt{x}+1)∣.$$
By our assumption that  $0<|x−1|<\delta$, we have
    $$|F(x) - L| <|1/(\sqrt{x}+1)|\delta = (1/(\sqrt{x}+1))ϵ(\sqrt{x}+1))  = \epsilon.$$
By letting $\delta=\epsilon(\sqrt{x}+1)$,  we get that $|x−1|<\epsilon$ if 0<|x−1|<δ.
Thus, $\lim_{x\to 1} F(x) = L$
The scratch work is usually omitted as far as finding the co-efficient of delta itself. Then just find the co-eff inverse and include epsilon and it falls out at the end. There is a good link on math exchange that shows a template of how to structure delta-epsilon proofs. It is how I learned to write them up, and this is that method. Good luck.
P.S. * represents multiplication, and the square roots were not working. I am unfamiliar with LaTex so this is the best I can do.
