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What I want to proof is

$\lim_{x\to 4}{\sqrt{x}}=2$

The following is from my textbook

Finding a $\delta$. Let $\epsilon\gt 0$. We seek a $\delta\gt 0$, s.t.

$if\quad 0\lt|x-4|\lt\delta,\quad then\quad|\sqrt{x}-2|\lt\epsilon$

To be able to form $\sqrt{x}$, we need to have $x\ge 0$. To ensure this, we must have $\delta\le 4$.

How to derive/translate this statement in inequality?

linear_combinatori_probabi
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    Without context, the question cannot be answered. – egreg Jan 15 '17 at 10:33
  • I will just point out that there are some older posts about this limit or very similar limits, which might also be useful for you. For example, http://math.stackexchange.com/questions/310611/epsilon-delta-proof-of-lim-x-to-4-sqrtx-2 or http://math.stackexchange.com/questions/2034743/real-analysis-prove-limit-lim-x-to-25-sqrt-x-5 – Martin Sleziak Jan 15 '17 at 17:51
  • @N1ng If you look at related and linked questions (shown in the sidebar on the right) for this question - and perhaps also recursively for some of related and linked questions - you can certainly find a few more. I suppose, some of them be more useful to you, some less, but they might be worth checking. – Martin Sleziak Jan 16 '17 at 04:04

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Note first that $|x-4|=|\sqrt x-2||\sqrt x+2|$ when $\sqrt x$ exists.

Now, let $\delta=\min(4,2\epsilon)$. Then if $|x-4|<\delta$, $\sqrt{x}$ exists (in particular, $|\sqrt x+2|\ge 2$), and

$$|\sqrt x-2|=\frac{|x-4|}{|\sqrt x+2|}\le\frac{1}{2}|x-4|<\frac{1}{2}(2\epsilon)=\epsilon.$$

Look at what has happened here. Why did we choose $\delta=\min(4,2\epsilon)$? We needed $\delta\le 4$ because this ensures that $\sqrt x$ exists, and we needed $\delta\le 2\epsilon$ to ensure that the inequality above worked out. Therefore the minimum of the two sufficed.

Edit: I will show that if $\delta\le 4$, then $|x-4|<\delta$ implies $x\ge0$.

Let $|x-4|<\delta$. If $x>4$, then of course $x>0$. Otherwise, $x\le 4$ and we have

$$4\ge\delta>|x-4|=4-x$$

which implies $0\ge -x$, i.e. $x\ge0$.

I think the way you should think of this intuitively is in terms of distances. That is, you should read $|x-4|<\delta$ as "the distance between $x$ and $4$ is smaller than $\delta$".

In particular, if $\delta\le 4$ then $|x-4|<\delta$ says "the distance between $x$ and $4$ is less than $4$", which means $x$ must be greater than $0$.

Alex Mathers
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