Suppose the functions $f$ and $g$ have the following property: for all $\varepsilon>0$ and all $x$, \begin{align}\text{if } 0< |x-2|&< \sin ^2{\left( \frac{\varepsilon ^2}{9}\right)} +\varepsilon \text{, then } |f(x)-2|<\varepsilon \\ \text{if } 0< |x-2|&<\varepsilon^2 \text{, then } |g(x)-4|<\varepsilon \end{align} For each $\varepsilon>0$ find a $\delta >0 $ such that, for all $x$,
(ii)$$\text{ if } 0<|x-2|<\delta\text{, then }|f(x)g(x)-8|<\varepsilon$$
I thought: $$|f(x)g(x)-8|=|f(x)g(x)-2g(x)+2g(x)-8|=|g(x)(f(x)-2)+2(g(x)-4)|$$ so with a choice of $\delta$ similar to this $\delta=\min\left(\sin ^2{\left( \frac{\varepsilon ^2}{9}\right)} +\varepsilon ,\varepsilon ^2/16 \right)$ and the triangle inequality, we can get: $$|g(x)(f(x)-2)+2(g(x)-4)|< \varepsilon |g(x)|+ \frac{\varepsilon}{2} $$
So what's left is to show that $|g(x)|\leq\frac{1}{2}$ for all $x$. But I can't figure out how to go about doing this and I'm not sure if there even is a way since the only way to get a bound for $g$ is from $-\varepsilon +4 \leq g(x) \leq \varepsilon+4$ and since $\varepsilon>0$ we can't get a tighter bound than $4 \leq g(x)\leq 4 \implies g(x)=4$. The solutions aren't very helpful either since the choice of $\delta$ there seems to be suggesting another way of going about it
We need
$$|f(x)-2|<\min\left( 1,\frac{\varepsilon}{2(|4|+1)}\right)\quad \text{and} \quad |g(x)-4|< \frac{\varepsilon}{2|2|+1} $$ so we need $$0<|x-2|<\min \left( \sin ^2 \left(\frac{[\min(1,\varepsilon/10)]^2]}{9} \right)+\min(1,\varepsilon /10), [\min(1,\varepsilon/6]^2\right)=\delta$$
And another question: if we had a function $h(x)=f(x)g(x)$ where $D_h=D_f \cap D_g$ then would my above method be sufficient to prove that $\displaystyle \lim_{x \to 2} h(x) =8 $? Since $\varepsilon |g(x)|+\varepsilon /2$ for all $x$ is of the form $C \varepsilon$ where $C\in \mathbb{R}^+$ and thus can get arbitrarily small. But in most cases where we have our $\varepsilon$ to be of that form $C\varepsilon$ at the end of the proof and we would like to get it to be $\varepsilon$, we can alter the $\varepsilon$ from the start to $\varepsilon /C$ but it seems like it isn't possible to do so here. Does that make any sense?
Rather than try to anticipate the correct factors you will need to get a desired quantity less than $\epsilon$, it is usually easier to show that you can get a desired quantity within $c\epsilon$ (for some $c$, say $c=100$ or $c=7$), then do a post-facto correction.
Step 1: Fix $\epsilon>0$. Motivated directly by the given info (without playing any preemptive scaling tricks) choose: $$ \delta = \min[\sin^2(\epsilon^2/9) + \epsilon, \epsilon^2]$$ Then for $x$ satisfying $0<|x-2|<\delta$ we get $$ |f(x)-2|<\epsilon, |g(x)-4|<\epsilon$$ and so by triangle inequality \begin{align} quantity=|g(x)(f(x)-2) + 2(g(x)-4)|&\leq |g(x)||f(x)-2|+2|g(x)-4| \\ &< |g(x)|\epsilon + 2\epsilon\\ &\leq (4+\epsilon)\epsilon + 2\epsilon \\ &=6\epsilon + \epsilon^2 \end{align}
Step 2: We notice if we assume $0<\epsilon\leq 1$ we can make $\epsilon^2\leq \epsilon$. So: Given any $\epsilon$ that satisfies $0<\epsilon\leq 1$ we can choose $\delta>0$ so that $0<|x-2|<\delta$ implies $$quantity < 7\epsilon$$
Step 3: So given any $\beta>0$, we can define $\epsilon = \min[\beta/7,1]$. So $0<\epsilon\leq 1$ and by step 2 we can choose $\delta>0$ so that $0<|x-2|<\delta$ implies $$quantity < 7\epsilon \leq \beta$$
The above is a basic way to reason through the problem. If you want you can then write the proof "backwards":
Fix $\beta>0$.
Define $\epsilon = \min[\beta/7, 1]$ [this is unmotivated to the reader].
Define $\delta = \min[sin^2(\epsilon^2/9)+\epsilon, \epsilon^2]$.
Show that $quantity< 7\epsilon \leq \beta$