Good evening, everybody! Some of you may remember my previous question regarding limits (I hope not, as my confusion was based on a pretty dumb error), and now I have another one. Again, the question I've chosen has been lifted verbatim from the 8th edition of Stewart Calculus (metric version), and again, I'm sorry for the lack of formatting. Anyway, on to the question.
Section 1.7, question 3: use the given graph of $f(x)=\sqrt{x}$ to find a number delta such that:
$$|x-4| < \delta\implies|\sqrt{x}-2| < 0.4$$
My answer: $\delta\leq1.44$. I got this by setting $f(x)$ equal to $1.6$ and $2.4$ and solving for the respective $x$'s, and choosing the lesser of the resulting $|x-4|$ expressions.
My point of confusion: working this out based on a graph is all well and good, but what if I don't have a graph or the tools to make one? Based on the section as a whole, there ought to be an algebraic process by which I can determine a maximum value for delta, right? Below is my attempt at such a process (be warned, it's not finished, as I got fed up and settled for the answer I got graphically in the first place.)
Let $f(x)=\sqrt{x}, a=4, L=2, \epsilon=0.4$. Find $\delta$ such that
$$0<|x-4|< \delta\implies|\sqrt{x}-2|<0.4$$
Now, the first thing I noticed is that $|x-4|$ can be rewritten as $|(\sqrt{x}-2)(\sqrt{x}+2)|$. Thus the first inequality becomes:
$$0<|\sqrt{x}-2||\sqrt{x}+2|< \delta$$
Setting $C=|\sqrt{x}+2|$:
if $0< C|\sqrt{x}-2|< \delta$, then $|\sqrt{x}-2|<0.4$
Getting rid of the absolute value signs, and rearranging:
$$(2-\delta)< C\sqrt{x}<(2+\delta)\implies 1.6< \sqrt{x}<2.4$$
Aaaaand that's as far as I got. Can anybody finish this line of reasoning for me, or explain where I've gone wrong? I appreciate any help; in the meantime, I'll continue to puzzle it over...
