The limit of a product of functions equals the product of the limits: Is this proof rigorous?
All the proofs I've seen so for the limit of a product of functions equaling the product of the limits are based on the following:
Let $f$ and $g$ be real or complex functions having the limits $$\lim_{x\to x_0}f(x) = F \quad \mbox{and} \quad \lim_{x\to x_0}g(x) = G.$$ Then also the limit $\displaystyle\lim_{x\to x_0}f(x)g(x)$ exists and equals $FG$.
Let $\varepsilon$ be any positive number.\, The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2,\,\delta_3$ such that \begin{align} |f(x)-F| < \frac{\varepsilon}{2(1+|G|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1} \end{align} \begin{align} |g(x)-G| < \frac{\varepsilon}{2(1+|F|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2} \end{align} \begin{align} |g(x)-G| < 1\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3\tag{3} \end{align} According to the condition (3) we see that $$|g(x)| = |g(x)\!-\!G\!+\!G| \leqq |g(x)\!-\!G|+|G| < 1\!+\!|G|\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3.$$ Supposing then that\, $0 < |x-x_0| < \min\{\delta_1,\,\delta_2,\,\delta_3\}$\, and using (1) and (2) we obtain \begin{align*} |f(x)g(x)-FG|\;& = |f(x)g(x)-Fg(x)+Fg(x)-FG|\\ & \leqq |f(x)g(x)\!-\!Fg(x)|+|Fg(x)\!-\!FG|\\ & = |g(x)|\cdot|f(x)\!-\!F|+|F|\cdot|g(x)\!-\!G|\\ & < (1\!+\!|G|)\frac{\varepsilon}{2(1\!+\!|G|)}+(1\!+\!|F|)\frac{\varepsilon}{2(1\!+\!|F|)}\\ & = \varepsilon \end{align*} This settles the proof.
But after having a go myself, I came up with the following:
Let $\varepsilon$ be any positive number. The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2$ such that \begin{align} |f(x)-F| < {\varepsilon}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1} \end{align} \begin{align} |g(x)-G| < {\varepsilon};\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2} \end{align}
Supposing then that\, $0 < |x-x_0| < \min\{\delta_1,\,\delta_2\}$\, and using (1) and (2) we obtain \begin{align*} |f(x)g(x)-FG|\;& = |(f(x) - F)(g(x) - G) + G(f(x) - F) + F(g(x) - G)|\\ & \leq |f(x) - F||g(x) - G| + |G||f(x) - F| + |F||g(x) - G|\\ & < \varepsilon^2 + \varepsilon(|L| + |G|)\\ & = \varepsilon' \end{align*}
where $\varepsilon'$ is any postive number, giving $\varepsilon = -1/2(|F|+|G|) +1/2\sqrt{(|F| + |G|)^2 + 4\varepsilon'}$
This settles the proof.
Is my proof rigorous enough?
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$\begingroup$Yes, the second proof is rigorous enough. But for clarity, it should add at the end:
Hence for any positive $\varepsilon'$, there exists a positive $\varepsilon < \varepsilon'$ that satisfies conditions (1) and (2).
This is important because you need $\varepsilon' \rightarrow 0 \Rightarrow\varepsilon \rightarrow 0$
$\endgroup$ $\begingroup$You can simplify it if you assume that $\varepsilon < 1$.
Then, from $|f(x)g(x)-FG| < \varepsilon^2 + \varepsilon(|F| + |G|) $ (you wrote $|L|$ instead of $|F|$) you get $|f(x)g(x)-FG| < \varepsilon(1+|F| + |G|) $ so you can just choose $\varepsilon \lt \dfrac{\min(1, \varepsilon')}{1+|F| + |G|} $.
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