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

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}})$$=\frac{\sqrt{3}+2}{100} and thus \vert{x-4}\vert<\frac{\sqrt{3}+2}{100} assures the case. The answer is actually \vert{x-4}\vert<\frac{1}{50}. I don't know where I went wrong because I took the steps needed in the proof using the epsilon-delta method. Was the restriction on \delta (\delta<1)wrong? I know the number 1 is arbitrary.. Does that mean there can be infinitely many conditions on \vert{x-a}\vert in general? ## 2 Answers You would like to find the tightest condition in this situation. For example if |x - 4| < \delta ensures the given condition, then any \delta' less than \delta also ensures the given condition. So, in order to do this you may like to find inf(\delta , \frac{\sqrt(4 - \delta) +2}{100}), and the \delta that minimizes that. So, this would we attained when the two terms are equal, which happens at \delta = \frac{199}{10000} < \frac{1}{50} . • Oh okay. I understand. But would there be any quicker way to work this question out? Jun 11, 2014 at 22:12 It would be a useful exercise, I would suggest, if you (the OP) write out explicitly the logical connections between your expressions, e.g.,$$|\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.

• Thank you for your suggestion! Jun 11, 2014 at 22:10
• @user143391, you're welcome. It can be very tricky, especially when you're first learning this stuff, to keep track of what it is you're assuming, and what you're trying to prove. Jun 11, 2014 at 22:14