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.
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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.
Best Answer
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 $2 > 1$.