# Math Help

1. Jun 6, 2005

### eutopia

our text book did not cover specifically this type of problem... can anyone help me?

Last edited by a moderator: May 2, 2017
2. Jun 6, 2005

### OlderDan

Notice that when x = 2, |x - 2| and |f(x) - 7| are both zero. If you graph those two absolute value functions you will see a "V" within a "V". Suppose $$\delta$$ is 1. For that value of $$\delta$$, x will have a maximum and minimum value that satisfies $$|x - 2| < \delta$$. If you use those limiting values of x, and look to the graph of |f(x) - 7| you will see that those values correspond to a certain value of |f(x) - 7|. That value is the value of $$\epsilon$$. For all values of x between those limiting values, $$|f(x) - 7| < \epsilon$$. You can find a relationship between $$\delta$$ and $$\epsilon$$ that shows they are proportional.

Last edited by a moderator: May 2, 2017
3. Jun 7, 2005

### eutopia

$$\epsilon$$ divided by 3 is what i got... i dont know if i did it right

4. Jun 7, 2005

### OlderDan

Looks good to me