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Prove $$\quad \lim_{x\to4}\sqrt{x}=2 $$

using the precise definition of limits. (Epsilon-Delta)

I think I proved this problem but when I look at the textbook to compare the proofs

they are quite different and I don't get how the book worked it all out.

So I am going to post mine for you to check if it's correct and the one from

the textbook to ask some questions. $\\[10pt]$

My rough work: $\;$ Pick $\epsilon > 0$, then there exists $\delta>0$ such that

$\quad 0 < |x - 4| < \delta \quad \Rightarrow \quad |\sqrt{x} - 2| < \epsilon$

Establish a connection between $|x - 4|$ and $|\sqrt{x} - 2|$

$|\sqrt{x} - 2| \cdot \frac{|\sqrt{x} + 2|}{|\sqrt{x} + 2|} = \frac{|x - 4|}{\sqrt{x} + 2}$, $\;$ Pick $\delta = 4$

$|x-4| < 4 \ \Rightarrow\ 0 < \sqrt{x} < \sqrt{8} \ \Rightarrow\ 2 < \sqrt{x} + 2 < \sqrt{8} + 2 \ \Rightarrow\ \frac{1}{\sqrt{8} + 2} < \frac{1}{\sqrt{x} + 2} < \frac{1}{2}$

This implies $\frac{|x - 4|}{\sqrt{x} + 2} < \frac{1}{2} \cdot |x-4| < \epsilon \ \Rightarrow\ |x-4|< 2 \cdot \epsilon$ $\\[20pt]$

My proof: $\;\delta = min\{4,2\epsilon\}$ and assume that $\ 0 < |x - 4| < \delta \ \Rightarrow \ |\sqrt{x} - 2| < \epsilon$

$\frac{|x - 4|}{\sqrt{x} + 2} < \frac{1}{2} \cdot |x-4| < \frac{1}{2} \cdot 2\epsilon < \epsilon \quad QED\quad $ Corrct? $\\[20pt]$

From the book: $\;$To be able to form $\sqrt{x}$, we need to have $x \ge 0$.

To ensure this, we must have $\delta \le 4$. With $x \ge 0$, we can form $\sqrt{x}$ and write

$|x - 4| = |\sqrt{x}+2||\sqrt{x}-2|$.

Since $|\sqrt{x} + 2| \ge 2 > 1$, it follows that $\quad \leftarrow\;$ This is what I don't understand.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $ where did "$\ge 2 > 1$" come from?

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $ the book doesn't explain

$|\sqrt{x}-2| \le |x - 4| \qquad \qquad \qquad \qquad \ \ \ \, \leftarrow$ How can I get this from the above inequality?

This last inequality suggests that we can simply set $\delta \le \epsilon \quad \leftarrow$ I got $2\epsilon$, how come it only has $\epsilon$?

Any help would be greatly appreciated.

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Probably is a typo. It should be $|\sqrt{x}+2|\geq 2>1$. Your proof is correct. –  P.. Feb 21 '13 at 22:16
    
@Pambos thanks, I corrected it. Can you explain where that inequality came from? –  Belphegor Feb 21 '13 at 22:21
    
Since $x\geq0\Rightarrow \sqrt x\geq0 \Rightarrow \sqrt{x}+2\geq0+2$. Note that in your proof (rough work) you have the same inequality! –  P.. Feb 21 '13 at 22:22
    
ohh i got it, then is 1 here important to note that $ \ge 2 > 1 $? –  Belphegor Feb 21 '13 at 22:27
    
To address your last question, setting $\delta \le \epsilon$ is okay, because remember that the whole game is, given $\epsilon$, find a candidate $\delta$. $\delta$ is not unique, we only need one that is good enough. You got $\delta = 2 \epsilon$, which is correct, because it is "good enough". Therefore, any smaller $\delta$ will also work. –  Christopher A. Wong Feb 21 '13 at 22:29

1 Answer 1

Answer to your where did $\geq2 > 1$ comes from:
It comes from the fact $\sqrt{x} \geq 0$, so $|\sqrt{x} + 2| \geq |0 + 2| = 2$ and of course $2 > 1$.

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