# 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

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!

• 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? Nov 19 '19 at 21:36
• This question really isn't identical. It's a more specific subquestion of the first. I don't have a problem with it. Nov 19 '19 at 21:37

## 2 Answers

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 $$\delta_1=\min\left(\frac{\sqrt{2}}{2},\frac{\epsilon}{2}\right)$$.

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

• 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. Nov 19 '19 at 22:00
• 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? Nov 20 '19 at 18:33
• You're right, see the edits. Nov 20 '19 at 19:49