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Help with question interpretation

  • Thread starter varignon
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  • #1
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1. Homework Statement

Give an example of a function f for which the following assertion is false:

If [tex]|f(x)-l|<\epsilon[/tex] when [tex]0<|x-a|<\delta[/tex], then [tex] |f(x)-l|<\epsilon/2[/tex], when [tex] 0<|x-a|<\delta/2[/tex]

3. The Attempt at a Solution

I am really not quite sure what I am looking for here. I think i want a function for which [tex]\delta[/tex] gets smaller much more quickly than epsilon does, any input as to what I am actually looking for would be great.
 

Answers and Replies

  • #2
HallsofIvy
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Since the problem only asks for "an example", I would go for the simplest. And it looks like a linear function, f(x)= mx+ b, should work. Draw an arbitrary straight line on an xy coordinate system, draw a rectangle at a point on that line, so the line is its diagonal, with [itex]\delta[/itex] as the length of the horizontal side and [itex]\epsilon[/itex] as the length of the vertical side. Now, imagine making [itex]\epsilon[/itex] smaller. How does [itex]\delta[/itex] change? What does the slope have to be so that [itex]\delta[/itex] decreases faster than [itex]\epsilon[/itex]?
 
  • #3
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If i let b =1, and if m < 1, then [tex]\delta[/tex] gets smaller more quickly than [tex]\epsilon[/tex]. Is f(x) = 0.25x + 1 a suitable answer to this question?
 
  • #4
HallsofIvy
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I may have answered too quickly and lead you astray. Yes, if m< 1, [itex]\delta[/itex] gets smaller more quickly than [itex]\epsilon[/itex]- but [itex]\delta[/itex] will reach half its original size exactly when [itex]\epsilon[/itex] reaches half its original size- so linear equations will not work here. Okay, then, what about y= x2? Take (0,0) as your initial point and [itex]\epsilon= 1[/itex]. What does [itex]\delta[/itex] have to be? Now take [itex]\epsilon= 1/2[/itex]. What does [itex]\delta[/itex] have to be.
 
  • #5
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[tex]\delta[/tex] has to be sqrt(2), which is still greater than [tex]\delta/2[/tex]. Further investigation revealed that this seemed to be the case for x^3 etc too. I tried it with 1/(x^2), as a=1, l=1, that seemed to work quite nicely. Is this a useful candidate?
 
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