Finding the Right Delta for Epsilon-Delta Proofs

[Paparazzi]

Homework Statement

Suppose the functions f and g satisfy the following property: for all \epsilon > 0 and all x, \text{if } 0 < | x - 2 | < \text{sin}^2(\frac{\epsilon^2}{9}) + \epsilon, \text{then } | f(x) - 2 | < \epsilon. \text{if } 0 < | x - 2 | < \epsilon^2, \text{then } | g(x) - 4 | < \epsilon.
For the given \epsilon > 0 find a \delta > 0 such that, for all x, \text{if } 0 < | x - 2 | < \delta, \text{then } | f(x) + g(x) - 6 | < \epsilon.

Homework Equations

N/A.

The Attempt at a Solution

Note that | f(x) + g(x) - 6 | = | (f(x) - 2) + (g(x) - 4) | \le | f(x) - 2 | + | g(x) - 4 | by the triangle inequality.
Now, we need to find \delta_1, \delta_2 > 0 such that the two conditions | f(x) - 2 | < \frac{\epsilon}{2} and | g(x) - 4 | < \frac{\epsilon}{2} are satisfied.

This is where I fall short of the solution. Since I've done this problem before I (unfortunately) know what the answer should be. It appears that I just substituted in \epsilon/2 for \epsilon in my previous solution, but I feel that it lacked rigor. Any clues as to where to go from here would be great. Thanks a lot.

 

[micromass]

Hi Paparazzi!

I appreciate your work to TeX everything nicely!

So, you need \delta_2 such that

0<|x-2|<\delta_2~\Rightarrow~|g(x)-4|<\epsilon/2

But, for all \epsilon, you have

0<|x-2|<\epsilon^2~\Rightarrow~|g(x)-4|<\epsilon

Thus if you make sure that \delta_2=(\epsilon/2)^2, then you will have

0<|x-2|<\delta_2=(\epsilon/2)^2~\Rightarrow~|g(x)-4|<\epsilon/2

Now you must do a similar thing for \delta_1...

 

[Paparazzi]

So since the statement is true for all \epsilon > 0, that's the reason you can just substitute in what is needed (since \epsilon/2 > 0)? Thanks a bunch for the reply.

 

[micromass]

So since the statement is true for all \epsilon > 0, that's the reason you can just substitute in what is needed (since \epsilon/2 > 0)?

Indeed!

 

[Paparazzi]

That is exactly what I needed. Thank you so much!