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Spivak Ch. 5 Limits, problem 6

  • Thread starter QuantumP7
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Homework Statement


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].


Homework Equations




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?
 

Answers and Replies

  • #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.
 
  • #3
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The denominator in the first line is 9, not g.
 
  • #4
jbunniii
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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?
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).
 
  • #5
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Thank you guys SO much!

jbunniii, that's exactly what I was looking for! Thank you!
 
  • #6
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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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