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Proof: $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$

Let $L_1 = \lim_{x \to a} f(x)$ and $L_2 = \lim_{x \to a} g(x)$. We assume $L_1, L_2 \neq 0$ (for scenarios where $L_1$ or $L_2$ are $0$, the result is trivially true).

Let $\epsilon > 0$. We need a $\delta$ such that $|f(x)g(x) - L_1L_2| < \epsilon$ whenever $0 < |x-a| < \delta$.

We can rewrite

$$\begin{align}|f(x)g(x) - L_1L_2| &= |(f(x)-L_1)(g(x)-L_2) + L_1(g(x)-L_2) + L_2(f(x)-L_1)| \\ &\leq |f(x)-L_1||g(x)-L_2| + |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\ &< \epsilon\end{align}$$

by triangle inequality.

Let $\delta_1 > 0$ such that $|f(x)-L_1| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_1$.

Let $\delta_2 > 0$ such that $|g(x)-L_2| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_2$.

Let $\delta_3 > 0$ such that $|g(x)-L_2| < \frac{\epsilon}{3|L_1|}$ whenever $0 < |x-a| < \delta_3$.

Let $\delta_4 > 0$ such that $|f(x)-L_1| < \frac{\epsilon}{3|L_2|}$ whenever $0 < |x-a| < \delta_4$.

Suppose that $0 < |x-a| < \delta$ where $\delta = \min(\delta_1, \delta_2, \delta_3, \delta_4)$. Then we have:

$$\begin{align}|f(x)-L_1||g(x)-L_2| &+ |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\&< \sqrt{\frac{\epsilon}{3}} \cdot \sqrt{\frac{\epsilon}{3}} + |L_1|\frac{\epsilon}{3|L_1|} + |L_2|\frac{\epsilon}{3|L_2|} \\&= \epsilon \end{align}$$

Is this a correct proof?

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your proof of the essential point seems flawless. one might only raise one small quibble: the statement of what you are proving is a little imprecise. suppose $f(x)=x$ and $g(x)=\frac1x$. then $\forall x.f(x)g(x)=1$ so

$$ \lim_{x \to 0} f(x)g(x) = 1 $$ but what is $\lim_{x\to 0}f(x) \lim_{x \to 0}g(x)$?

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  • $\begingroup$ I'm a little confused, does this make the entire proof wrong? Do I have to account for some extra case? What am I missing exactly? Accounting for infinity separately? Indeterminate forms? $\endgroup$
    – user525966
    Mar 1, 2018 at 1:45
  • $\begingroup$ not at all, it is a very nice proof. you can try formulations like (abbreviated here) "if Limit f and Limit g exist, then Limit fg exists and Limit fg = Limit f * Limit g" $\endgroup$ Mar 1, 2018 at 1:49
  • $\begingroup$ your proof shows precisely that if the two limits exist then the product of the functions also has a limit and it is the product of the limits. but what you want to avoid is the implication that if the product of two functions has a limit then the functions separately each have limits. it is just a matter of careful phrasing. $\endgroup$ Mar 1, 2018 at 2:54
  • $\begingroup$ What do I need to add to make the proof more complete? $\endgroup$
    – user525966
    Mar 1, 2018 at 2:58

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