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Homework Statement
Ok so this may get a little drawn out here because my book only gives me one example and I guess I can't decipher its meaning. So here is the example they give:
For \;\; \lim_{x \to 2} x^{2} = 4
Find a \;\; \delta > 0 \;\; such that whenever
0 < |x-2|< \delta, \;\;\; |x^{2}-4| < \frac {1}{10}
By the Lemma, we must find \delta > 0 such that whenever
2-\delta < x < 2+ \delta \;\;\;\; and\;\;\;\; x\neq2
4-\frac{1}{10} < x^{2} <4+\frac{1}{10}
(note that the only thing that the lemma states that that sentence is referring to is defining within delta and strictly within delta)
Assume that
2-\delta < x \;\;\;and\;\;\; x < 2+\delta
As long as \;\;\;2-\delta\;\;\; and \;\;\;x\;\;\; are positive we may square both sides and get:
4-4\delta+\delta^{2} < x^{2} \;\;\;\; and \;\;\;\; x^{2} < 4+4\delta+\delta^{2}
4+(-4\delta+\delta^{2}) < x^{2} \;\;\;\; and \;\;\;\; x^{2} < 4+(4\delta+\delta^{2})
(So here is question one, why did they square the inequality? Is it because f(x)=x^2 or some other reason that I am not picking up? This is where I would better understand what was going on with more than example being given)
Now take a \;\; \delta \;\; small enough so that:
-\frac {1}{10} \leq -4\delta + \delta^{2} \;\;\;\; and \;\;\;\; 4\delta+\delta^{2} \leq \frac {1}{10}
(So question two, why did they take only the \; -4\delta+\delta^{2} \; etc out instead of the entire left/right side? Is this only applicable to this problem and the squaring or does something similar happen on other epsilon, delta condition problems?)
Use 1/50 for example delta value and thus:
0<|x-2|<\frac {1}{50}, \;\;\; |x^{2}-4| < \frac {1}{10}
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So an actual problem given at the end of the chapter is this:
For:
\lim_{x \to 2} \frac {1}{x} = \frac {1}{2}
Find a \; \delta > 0 \; such that whenever:
0 < |x-2| < \delta, \;\;\; |\frac{1}{x}-\frac{1}{2}| < .01
and the lemma (referenced above) states that x is strictly within \; \delta \; of c if and only if:
c-\delta<x<c+\delta
and in this problem c = 2 so this must be true:
2-\delta<x<2+\delta
so at this point I could make x into f(x) by doing a reciprocal like:
\frac{1}{2-\delta}<\frac{1}{x}<\frac{1}{2+\delta}
and if that is correct the next step i think would be this:
-.01 \leq \frac {1}{2-\delta} \;\;\;\; and \;\;\;\; \frac{1}{2+\delta} \leq .01
and if that is all correct I would just have to find a delta that is true for both those inequalities. I will stop here though so that I can get a yes no to my method. If this is way off then I apologize.
Thanks for any help!