Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$–$\delta$ definition of limit.
Method 1
Let $\epsilon>0$
Choose $\delta<\epsilon-6$. ($\delta>0$)
If $|x-4|<\delta \implies -\delta < x-4 < \delta \implies 4-\delta<x<4+\delta$,
then, $|\sqrt{x}-2| < \sqrt{x}+2 < \sqrt{4+\delta}+2 < 6+\delta < \epsilon$
Hence, by $\epsilon$–$\delta$ definition of limit, $\lim_{x\to 4} \sqrt{x} = 2$
Is method 1 correct?
Method 2
Let $\epsilon>0$
Choose $\delta<$___. ($\delta>0$)
If $|x-4|<\delta \implies -\delta < x-4 < \delta \implies 4-\delta<x<4+\delta$,
then, $|\sqrt{x}-2| < \sqrt{x}+2 < \sqrt{4+\delta}+2$
(Scratch work: $\sqrt{4+\delta}+2=\epsilon \implies \delta+4=(\epsilon-2)^{2} \implies \delta=\epsilon^{2}-4\epsilon$.)
Now I'm tempted to fill the blank by choosing $\delta<\epsilon^{2}-4\epsilon$. But when I do that I'm not getting the desired $|\sqrt{x}-2| < \epsilon$ because $\sqrt{4+\delta}+2<\sqrt{4+\epsilon^{2}-4\epsilon}+2=\sqrt{(\epsilon-2)^{2}}+2=|\epsilon-2|+2$, which is not necessarily less than $\epsilon$.
What's wrong in method 2?
Best Answer
Scratch Work
The goal is to determine a value of $\delta > 0$ such that $|x-4| < \delta$ implies that $|\sqrt{x} - 2| < \varepsilon$. Note
$$ \bigl| x-4 \bigr| = \bigl| (\sqrt{x}-2)(\sqrt{x}+2) \bigr|. \label{eq1}\tag{*}$$
Observe that $| x-4 | < \delta$ implies \begin{align} -\delta < x-4 < \delta &\iff 4-\delta < x < 4+\delta \\ &\iff \sqrt{4-\delta} < \sqrt{x} < \sqrt{4+\delta} \\ &\iff \sqrt{4-\delta}+2 < \sqrt{x}+2 < \sqrt{4+\delta}+2. \end{align} Hence $\sqrt{x}+2$ it is possible to conclude that $\sqrt{x}+2 > 2$ for any value of $\delta$ such that $\sqrt{4-\delta}$ "makes sense", i.e. $\delta < 4$. This provides some control over the second factor in (\ref{eq1}) (it will be bigger than $1$, which is helpful). The other factor is $\sqrt{x}-2$, which is the thing we are trying to make small. In particular, $$\bigl| \sqrt{x}-2 \bigr| < \bigl| (\sqrt{x}-2)(\sqrt{x}+2) \bigr| = \bigl| x-4\bigr| < \delta. $$ Of particular note is that $|\sqrt{x}-2|$ is now bounded by $\delta$, which is exactly what is desired, as long as $\delta < \varepsilon$.
So choose $\delta$ to be smaller than $4$ (to bound $\sqrt{x}+2$), and smaller than $\varepsilon$ (to get the desired result in the end).
Proof
Fix $\varepsilon > 0$ and choose $\delta < \min\{4, \varepsilon\}$. Suppose that $|x-4| < \delta$. Since $\delta < 4$, if $|x-4|<\delta$, then $$ |x-4| < 4 \implies 0 < x \implies 2 < \sqrt{x}+2. $$ In particular, $\sqrt{x}+2 > 1$, hence \begin{align} \bigl|\sqrt{x} - 2\bigr| &< \bigl| (\sqrt{x}+2)(\sqrt{x}-2)\bigr| && \text{(since $\sqrt{x}+2 > 1$)} \\ &= \bigl| x-4 \bigr| \\ &< \delta \\ &< \varepsilon, \end{align} which is the desired result.