Determine a condition on $\vert{x-4}\vert$ that will assure...

Question

This is a real analysis question regarding limits. The question is to determine a condition on $\vert{x-4}\vert$ that will assure that $\vert{\sqrt{x}-2}\vert<\frac{1}{100}$.

My work:

$\vert{\sqrt{x}-2}\vert<\frac{1}{100}$

$\frac{\vert{\sqrt{x}-2}\vert\vert{\sqrt{x}+2}\vert}{\vert{\sqrt{x}+2}\vert}<\frac{1}{100}$

$\frac{\vert{x-4}\vert}{\vert{\sqrt{x}+2}\vert}<\frac{1}{100}$

$\vert{x-4}\vert<\frac{\vert{\sqrt{x}+2}\vert}{100}$

If we put a restriction on $\delta$ s.t. $\delta<1$ Then, $\vert{x-4}\vert<1$. Thus, $3<x<5$ and $\sqrt{3}+2<\sqrt{x}+2<\sqrt{5}+2$

So,$\vert{x-4}\vert<\frac{\sqrt{3}+2}{100}$

$\delta=\inf({1,\frac{\sqrt{3}+2}{100}})$$=\frac{\sqrt{3}+2}{100}$ and thus $\vert{x-4}\vert<\frac{\sqrt{3}+2}{100}$ assures the case.

The answer is actually $\vert{x-4}\vert<\frac{1}{50}$. I don't know where I went wrong because I took the steps needed in the proof using the epsilon-delta method. Was the restriction on $\delta$ ($\delta<1$)wrong? I know the number 1 is arbitrary.. Does that mean there can be infinitely many conditions on $\vert{x-a}\vert$ in general?

Answer 1

You would like to find the tightest condition in this situation. For example if $|x - 4| < \delta$ ensures the given condition, then any $\delta'$ less than $\delta$ also ensures the given condition. So, in order to do this you may like to find $inf(\delta , \frac{\sqrt(4 - \delta) +2}{100})$, and the $\delta$ that minimizes that. So, this would we attained when the two terms are equal, which happens at $\delta = \frac{199}{10000} < \frac{1}{50} $.

Answer 2

It would be a useful exercise, I would suggest, if you (the OP) write out explicitly the logical connections between your expressions, e.g.,

$$|\sqrt x-2|\lt{1\over100} \implies{|\sqrt x-2||\sqrt x+2|\over|\sqrt x+2|}\lt{1\over100} \implies{|x-4|\over|\sqrt x+2|}\lt{1\over100} \implies\text{etc.}$$

and then ask yourself which $\implies$ signs can be replaced with the if-and-only sign, $\iff$. I think you'll find that somewhere the line, one of them cannot.