This is problem 1.7 44b from Calculus 7th Edition by James Stewart. The problem asks me to write an $\varepsilon$-$\delta$ proof on the limit $$\lim_{x\to a}[f(x)g(x)] = \infty$$ Given that $$\lim_{x\to a}f(x) = \infty\qquad and\qquad \lim_{x\to a}g(x) = c\quad where\quad c > 0$$
My solution:
Suppose that
(1) for every $M_1>0$, there is a $\delta_1>0$ such that if $0<|x-a|<\delta_1$ then $f(x)>M_1$
(2) for every $\varepsilon>0$, there is $\delta_2>0$ such that if $0<|x-a|<\delta_1$ then $|g(x)-c|<\varepsilon$.
I want to show that for every $M_2>0$, there is $\delta_3>0$ such that if $0<|x-a|<\delta_3$ then $f(x)g(x) > M_2$.
To satisfy the two assumptions, I guess $\delta_3 = min(\delta_1,\delta_2)$.
Proof
Given a positive number $M_2$ I choose $\delta_3 = min(\delta_1,\delta_2)$, so that
(1) if $0<|x-a|<\delta_3$ then $f(x)>M_1$
(2) if $0<|x-a|<\delta_3$ then $|g(x)-c|<\varepsilon$ where $c>0$.
I then start by rearranging inequality (2).
$$-\varepsilon<g(x)-c<\varepsilon$$
$$-\varepsilon+c<g(x)<\varepsilon+c$$
Since $M_1>0$ and $f(x)>0$, then multiplying this with the inequality above gives
$$f(x)g(x) > M_1(-\varepsilon+c)$$
Now I know that to continue the proof, I must show that $M_1(-\varepsilon+c)$ is positive and it can be as large as one please; however, the problem is $(-\varepsilon+c)$ could be negative which makes $M_1(-\varepsilon+c)$ negative because $M_1>0$. I know that I must assume $0<\varepsilon<c$ but I don't see the reason why I can make such assumption since according to the definition, all $\varepsilon>0$ must work. Can anyone please explain why? Thank you very much.
$\epsilon$ is completely your choice. Your are given that $\lim_{x \to a} g(x) =c$ and you can apply the definition of limit here with any $\epsilon $ you like. So choose any $\epsilon \in (0,c)$, say $\epsilon =\frac c2$, and your proof works fine.