Use the epsilon-delta definition to show that $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$ $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$
by using the epsilon-delta method.
I have been trying so solve this for the last hour, but I'm completely stuck.
Help, anyone?
 A: Fix $\epsilon > 0$.
Find $\delta_1$ such that $|\frac{2}{x} - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x-\sqrt{2}| < \delta_1$.
Find $\delta_2$ such that $|x - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x - \sqrt{2}| < \delta_2$.
Then let $\delta = \min\{\delta_1, \delta_2\}$.

 We then have $$|\frac{1}{2} (\frac{2}{x} + x) - \sqrt{2}| \le \frac{1}{2} |\frac{2}{x} - \sqrt{2}| + \frac{1}{2} |x - \sqrt{2}| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ for any $x$ satisfying $|x - \sqrt{2}| < \delta$.


Response to comment:
If $|x| \ge 1$, then
$$|\frac{2}{x} - \sqrt{2}|
= \sqrt{2} \left| \frac{\sqrt{2} - x}{x}\right|
\le \sqrt{2} |\sqrt{2} - x| < \sqrt{2} \delta_1.$$
So, choose $\delta_1 = \min\{\sqrt{2}-1, \epsilon/\sqrt{2}\}$.
(The $\sqrt{2}-1$ is to ensure $|\sqrt{2}-x|<\delta_1 \le \sqrt{2}-1 \implies |x| \ge 1$.)
A: $\begin{array}\\
\frac{1}{2}(\frac{2}{x}+x) - \sqrt2
&=\frac{1}{2}(\frac{2}{x}-2\sqrt{2}+x)\\
&=\frac{1}{2}(\sqrt{\frac{2}{x}}-\sqrt{x})^2\\
&=\frac{1}{2}(\frac1{\sqrt{x}}(\sqrt{2}-x))^2\\
&=\frac{1}{2x}(\sqrt{2}-x)^2\\
\end{array}
$
This makes it clear
what happens as
$x \to \sqrt{2}$.
