# 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? – Robert Shore 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. – Robert Shore Nov 19 '19 at 21:37

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 $$\epsilon<\sqrt{2}$$
• 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? – Sondr Nov 20 '19 at 18:33