Suppose the functions $f$ and $g$ have the following property: for all $\epsilon > 0$ and all x :
if $0 < |x-2| < \sin^2(\frac{\epsilon ^2}{9}) + \epsilon \Rightarrow |f(x) - 2| < \epsilon$ and if $0 < |x-2| <\epsilon ^2 \Rightarrow |g(x) - 4| < \epsilon$.
For each $\epsilon > 0$, find a $\delta > 0$ such that:
if $0 < |x-2| < \delta$ then $|f(x)g(x) - 8|< \epsilon$
Attempt: I was trying to manipulate the expression $|f(x)g(x) - 8|$ with the hope of getting it into the form $|f(x) - 2||g(x) - 4|$ where I could draw some conclusions from the given statements, but I was unsuccessful in these attempts. I saw the solution and it states:
we need: $$|f(x) - 2| < \min \left(1, \frac{\epsilon}{2(|4| + 1)}\right)$$ and $$|g(x) - 4| < \frac{\epsilon}{2(|2| + 1)}$$
And from this we're able to draw a conclusion on $\delta$. My question is how did they go about obtaining these values and the train of thought used to conclude this? I've been doing delta-epsilon proofs for ahwile and never saw this process before.
Edit Ok I've been working on it and I almost have the solution I believe:
I was hinted to bound g(x), so attempting that:
$$\text{let } |g(x) - 4| < 4 \Rightarrow 0 < g(x) < 8$$
Therefore: $$|8||f(x) - 2| + |2||g(x) - 4| = \\ |4||f(x)-2| + |g(x) -4| < \frac{\epsilon}{2}$$
I can see how $(|4| + 1)$ can come about, but in order for that to occur, I would have needed $|g(x) - 4| = |f(x) - 2|$.....suggesstions?
Hint: $$f(x)g(x)-8=f(x)g(x)-2g(x)+2g(x)-8=(f(x)-2)g(x)+2(g(x)-4).$$
Then use triangle inequality and you may want to bound $|g(x)|$ from above using the fact that the limit of $g$ is 4.