Use the epsilon-delta definition to show that $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$

Question

$\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$

by using the epsilon-delta method.

I have been trying so solve this for the last hour, but I'm completely stuck.

Help, anyone?

Answer 1

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 $|x - \sqrt{2}| < \delta$.

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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 $|\sqrt{2}-x|<\delta_1 \le \sqrt{2}-1 \implies |x| \ge 1$.)

Answer 2

$\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 $x \to \sqrt{2}$.