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?
Best Answer
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 $quantity< 7\epsilon \leq \beta$