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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

  1. Find $\delta_1 > 0$ such that $|x-\sqrt2|<\delta_1$ implies $|\frac{2}{x}-\sqrt2| < \epsilon$ .
  2. Find $\delta_2 > 0$ such that $|x-\sqrt2|<\delta_2$ implies $|x-\sqrt2| < \epsilon$ .

  3. 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!

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  • $\begingroup$ What you're trying to do is show that $\lim_{x \to \sqrt 2} \frac 2x = \sqrt 2$. More generally, you're trying to show that $\lim_{x \to a} \frac 2x = \frac 2a$. Do you know how to do that? $\endgroup$ – Robert Shore Nov 19 '19 at 21:36
  • $\begingroup$ This question really isn't identical. It's a more specific subquestion of the first. I don't have a problem with it. $\endgroup$ – Robert Shore Nov 19 '19 at 21:37
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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 $\delta_1=\min\left(\frac{\sqrt{2}}{2},\frac{\epsilon}{2}\right)$.

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$|\frac{2}{x}-\sqrt2| < \epsilon\Leftrightarrow -\epsilon<\frac{2}{x}-\sqrt2 < \epsilon\Leftrightarrow -\epsilon+\sqrt2<\frac{2}{x} < \epsilon+\sqrt2\Leftrightarrow \frac{1}{\sqrt2-\epsilon}>\frac{x}{2} > \frac{1}{\sqrt2+\epsilon}\Leftrightarrow \frac{2}{\sqrt2-\epsilon}>x > \frac{2}{\sqrt2+\epsilon}\Leftrightarrow \frac{2}{\sqrt2-\epsilon}-\sqrt{2}>x-\sqrt{2} > \frac{2}{\sqrt2+\epsilon}-\sqrt{2} \Leftarrow\left|x-\sqrt{2}\right|<\min\left(\frac{2}{\sqrt2-\epsilon}-\sqrt{2},\sqrt{2}-\frac{2}{\sqrt2+\epsilon}\right)=\delta_1$ The third "$\Leftrightarrow$" holds for small enough $\epsilon<\sqrt{2}$

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  • $\begingroup$ Formatting nit-pick: that long chain of relations should be split into several lines, rather than depending on the MathJax renderer to do that automatically, because some renderers don't. $\endgroup$ – Calum Gilhooley Nov 19 '19 at 22:00
  • $\begingroup$ Correct me if I'm wrong, but since $\epsilon > 0$, then the last $\Leftarrow$-implication $\left|x-\sqrt{2}\right|<\min\left(\frac{2}{\sqrt2-\epsilon}-\sqrt{2},\frac{2}{\sqrt2+\epsilon}-\sqrt{2}\right)=\delta_1$ states that $\delta_1 < 0$ (while it should me positive) and also that somehow $\left|x-\sqrt{2}\right| < \delta_1 < 0$ . Could you review your answer? $\endgroup$ – Sondr Nov 20 '19 at 18:33
  • $\begingroup$ You're right, see the edits. $\endgroup$ – Alexey Burdin Nov 20 '19 at 19:49

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