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

How would I prove this? It is an example in the textbook, but I am pretty confused.

Martin Sleziak
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Mat
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1 Answers1

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Let $\epsilon > 0$. You want to find $\delta > 0$ such that $|x-4|<\delta$ implies $|\sqrt{x} - 2| < \epsilon$. This is merely the defintion of limit.

Do note that: $$|\sqrt{x}-2| = |\sqrt{x}-2|\frac{|\sqrt{x}+2|}{|\sqrt{x}+2|} = \frac{|x-4|}{|\sqrt{x}+2|} < \frac{\delta}{|\sqrt{x}+2|} \color{red}{< \epsilon}.$$We must find some condition on $|\sqrt{x}+2|$ to get that last red inequality that we want. Here this is simple, because: $$|\sqrt{x}+2| = \sqrt{x}+2 \geq 2 > 0 \implies \frac{1}{|\sqrt{x}+2|} \leq \frac{1}{2},$$so that $\delta < 2\epsilon$ will work.

You can read my answer here, in which I explain things more thoroughly.

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Ivo Terek
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    You should suppose $\delta \le 1$, not $\delta<1$, because you later write $\min{1,(2-\sqrt{3})\epsilon}$. I've edited this. – user236182 Oct 12 '15 at 22:37
  • How did you go from |x-4| < δ < 1 to |x| > 3. I know when you simplify the first part you get 3 < x < 5 – Mat Oct 12 '15 at 22:59
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    Using that $|x| - 4 \leq ||x|-|4|| \leq |x-4| < 1$. – Ivo Terek Oct 12 '15 at 23:00
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    Although looking carefully, this whole talk of picking $\delta \leq 1$ is unnecessary here, I'll edit my answer to simplify it. Although you should really read my other answer to learn that trick, it is very useful. – Ivo Terek Oct 12 '15 at 23:01
  • Doesn't that give you |x| < 5 ? – Mat Oct 12 '15 at 23:05
  • Yes. We get $3 < |x| < 5$. $$|x| - 4 < 1 \implies |x| < 5, \quad 4 - |x| < 1 \implies |x| > 3.$$ – Ivo Terek Oct 12 '15 at 23:08