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!