Help with question interpretation

[varignon]

Homework Statement

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

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

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 \delta gets smaller much more quickly than epsilon does, any input as to what I am actually looking for would be great.

 

[HallsofIvy]

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 \delta as the length of the horizontal side and \epsilon as the length of the vertical side. Now, imagine making \epsilon smaller. How does \delta change? What does the slope have to be so that \delta decreases faster than \epsilon?

 

[varignon]

If i let b =1, and if m < 1, then \delta gets smaller more quickly than \epsilon. Is f(x) = 0.25x + 1 a suitable answer to this question?

 

[HallsofIvy]

I may have answered too quickly and lead you astray. Yes, if m< 1, \delta gets smaller more quickly than \epsilon- but \delta will reach half its original size exactly when \epsilon 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 \epsilon= 1. What does \delta have to be? Now take \epsilon= 1/2. What does \delta have to be.

 

[varignon]

\delta has to be sqrt(2), which is still greater than \delta/2. 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?