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Spivak Ch. 5 Limits, problem 6

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose the functions f and g have the following property: for all ε > 0 and all x,

    If [itex]0 < \left| x-2 \right| < \sin^{2} \left( \frac{\varepsilon^{2}}{9} \right) + \varepsilon[/itex], then [itex]\left| f(x) - 2 \right| < \varepsilon[/itex].

    If [itex]0 < \left| x - 2 \right| < \varepsilon^{2}[/itex] then [itex]\left| g(x) - 4 \right| < \varepsilon[/itex].

    For each [itex]\varepsilon > 0[/itex] find a [itex]\delta > 0[/itex] such that for all x:


    ii) if [itex]0 < \left| x-2 \right| < \delta[/itex], then [itex]\left| f(x)g(x) - 8 \right| < \varepsilon[/itex].


    2. Relevant equations


    3. The attempt at a solution

    The solution book says:

    We need: [itex]\left| f(x) - 2 \right| < min \left( 1, \frac{\varepsilon}{2 (\left| 4 \right| + 1)} \right) [/itex] and [itex]\left| g(x) - 4 \right| < \frac{\varepsilon} {2 (\left| 2 \right|) + 1}[/itex].

    My question is: How did they get these fractions in the solution? I've multiplied [itex]\left| f(x) - 2 \right|[/itex] and [itex]\left| g(x) - 4 \right| [/itex] to get [itex]\left| f(x)g(x) - 4f(x) - 2g(x) + 8 \right|[/itex]. Am I going in the right direction?
     
  2. jcsd
  3. Jan 23, 2013 #2

    haruspex

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    That g in the denominator in the first line is g(x)?
    I would start by making some simplifying substitutions: u = x+2, F(u) = f(x)-2, G(u) = g(x)-4.
     
  4. Jan 23, 2013 #3
    The denominator in the first line is 9, not g.
     
  5. Jan 24, 2013 #4

    jbunniii

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    You'll have to do some more algebra to get it into a useful form. Here's a suggestion for how to start:
    $$f(x)g(x) - 8 = (f(x) - 2)(g(x) - 4) + 4f(x) + 2g(x) - 16$$
    Now see if you can do some more manipulations on the right hand side. You want to get more terms containing (f(x) - 2) and (g(x) - 4).
     
  6. Jan 24, 2013 #5
    Thank you guys SO much!

    jbunniii, that's exactly what I was looking for! Thank you!
     
  7. Sep 24, 2013 #6
    A note for future visitors: the solution follows directly from part (2) of the lemma. No need for the algebraic manipulations.
     
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