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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?

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2 Answers 2

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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} $.

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  • $\begingroup$ Oh okay. I understand. But would there be any quicker way to work this question out? $\endgroup$
    – user143391
    Jun 11, 2014 at 22:12
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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.

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  • $\begingroup$ Thank you for your suggestion! $\endgroup$
    – user143391
    Jun 11, 2014 at 22:10
  • $\begingroup$ @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. $\endgroup$ Jun 11, 2014 at 22:14

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