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?