Use the $\epsilon$ - $\delta$ definition to show that $\lim_{x\to \sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$ Use the epsilon-delta definition to show that $\lim_{x\to \sqrt2} \frac{1}{2}\left(\frac{2}{x}+x\right) = \sqrt2$.
I have been shown the following approach to solve this:

Let first $\epsilon > 0$.
Then
  
  
*
  
*Find $\delta_1 > 0$ such that $|x-\sqrt2|<\delta_1$ implies $|\frac{2}{x}-\sqrt2| < \epsilon$ .
  
*Find $\delta_2 > 0$ such that $|x-\sqrt2|<\delta_2$ implies $|x-\sqrt2| < \epsilon$ .
  
*Then let $\delta = \min\{\delta_1,\delta_2\}$ .
One would then have
  $$\left\lvert\frac{1}{2} \left(\frac{2}{x} + x\right) - \sqrt{2}\right\rvert| \le \frac{1}{2} \left\lvert\frac{2}{x} - \sqrt{2}\right\rvert + \frac{1}{2} |x - \sqrt{2}| < \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon$$ for any $x$ satisfying $|x - \sqrt{2}| < \delta$.

Ok, so working my way backwards through this.
I understand the last step if points 1), 2), and 3) have been done.
I understand why 3) is done.
I understand that in 2) one can simply set $\delta_2 = \epsilon$ .
What I don't get, is how you find $\delta_1$ as described in 1).
I have little experience with epsilon-delta proofs/verification. 
I appreciate any help I can get!
 A: First of all, you don't have to prove it this way. You might try to directly work with $\left|\frac{1}{2}\left(\frac{2}{x}+x\right)-\sqrt{2}\right|$, as shown here in the first question you posted about this problem. But if you want to work this way, then
$$\left|\frac 2x-\sqrt{2}\right|=\sqrt{2}\left|\frac{\sqrt{2}}{x}-1\right|=\sqrt{2}\left|\frac{\sqrt{2}-x}{x}\right|=\sqrt{2}\frac{|x-\sqrt{2}|}{|x|}$$
Now you need to find an upper bound of $\frac{1}{|x|}$. Assuming $|x-\sqrt{2}|<\frac{\sqrt{2}}{2}$, you have $x\in\left(\frac{\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right)$, so $|x|>\frac{\sqrt{2}}{2} \Rightarrow \frac{1}{|x|}<\frac{2}{\sqrt{2}}$, which implies that
$$\sqrt{2}\frac{|x-\sqrt{2}|}{|x|}<\sqrt{2}\cdot\frac{2}{\sqrt{2}}|x-\sqrt{2}|=2|x-\sqrt{2}|$$
To ensure this is less than $\epsilon$, you need $|x-\sqrt{2}|<\frac{\epsilon}{2}$, i.e., you choose $\delta_1=\min\left(\frac{\sqrt{2}}{2},\frac{\epsilon}{2}\right)$.
A: $|\frac{2}{x}-\sqrt2| < \epsilon\Leftrightarrow
-\epsilon<\frac{2}{x}-\sqrt2 < \epsilon\Leftrightarrow
-\epsilon+\sqrt2<\frac{2}{x} < \epsilon+\sqrt2\Leftrightarrow
\frac{1}{\sqrt2-\epsilon}>\frac{x}{2} > \frac{1}{\sqrt2+\epsilon}\Leftrightarrow
\frac{2}{\sqrt2-\epsilon}>x > \frac{2}{\sqrt2+\epsilon}\Leftrightarrow
\frac{2}{\sqrt2-\epsilon}-\sqrt{2}>x-\sqrt{2} > \frac{2}{\sqrt2+\epsilon}-\sqrt{2}
\Leftarrow\left|x-\sqrt{2}\right|<\min\left(\frac{2}{\sqrt2-\epsilon}-\sqrt{2},\sqrt{2}-\frac{2}{\sqrt2+\epsilon}\right)=\delta_1$
The third "$\Leftrightarrow$" holds for small enough $\epsilon<\sqrt{2}$
