Proof writing: $\varepsilon$-$\delta$ proof of the product rule for limits — any tips for cleaning this up

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I've written the following proof for the product rule for limits. Here $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$. While it works, I find it to be clunky and I can't shake the feeling that we somehow can reduce the general case to the much (in my opinion) cleaner proof of the special case where $L = M = 0$. Anyhow, here is the proof:

Suppose first that $L = M = 0$. Fix an arbitrary $\varepsilon > 0$ and let $\delta_1$ and $\delta_2$ be such that $f(x)$ and $g(x)$ are in a $\sqrt{\varepsilon}$ neighborhood of $0$, respectively. Choose $\delta = \min(\delta_1, \delta_2)$. Then when $|x-c| < \delta$ we have
$$ |f(x)g(x)| = |f(x)||g(x)| < \left(\sqrt{\varepsilon}\right)^2 = \varepsilon. $$

Suppose instead $M \neq 0$. Fix an arbitrary $\varepsilon > 0$ and let $\delta_1$ be such that it satisfies $|f(x) – L| < \varepsilon_1 = \frac{\varepsilon}{3|M|}$. Moreover, let
$\delta_2$ be such that it satisfies $|g(x) – M| < \varepsilon_2 =
\min(\frac{\varepsilon}{3\varepsilon_1}, \frac{\varepsilon}{3|L|})$ where
$\frac{\varepsilon}{3|L|} = \infty$ if $L = 0$. Choose $\delta = \min(\delta_1, \delta_2)$. Then if $|x – c| < \delta$, we have
\begin{align*}
|f(x)g(x) – LM| &= |f(x)g(x) – Lg(x) + Lg(x) – LM| \\
&= |g(x)(f(x) – L) + L(g(x) – M)| \\
&\leq |g(x)||(f(x) – L)| + |L||(g(x) – M)| \\
&\leq |M \pm \varepsilon_2|\varepsilon_1 + |L|\varepsilon_2 \\
&\leq |M|\varepsilon_1 + \varepsilon_1\varepsilon_2 + |L|\varepsilon_2 \\
&\leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} < \varepsilon.
\end{align*}

Best Answer

Sure, the general case can be reduced to the case $L = M = 0$ by subtracting $L$ and $M$ from $f$ and $g$, respectively. Once we know that $\lim_{x\to c}f(x)g(x) = 0$ whenever $f(x),g(x)\to 0$ as $x\to c$, we deduce $$\lim_{x\to c}(f(x)-L)(g(x)-M) = 0.$$ Using algebraic limit laws, this simplifies to $$\lim_{x\to c}f(x)g(x) = LM.$$

Added: The algebraic limit laws I have in mind are (here all the limits are presumed to exist).

  • $\displaystyle\lim_{x\to c}[F(x) + G(x)] = \lim_{x\to c}F(x) + \lim_{x\to c}G(x)$.
  • For any real $\alpha$, $\displaystyle\lim_{x\to c}\alpha F(x) = \alpha\lim_{x\to c}F(x)$.

As for cleaning up the direct proof in the general case, I would proceed like this.

Proof. By the triangle inequality, $$|f(x)g(x)-LM| \le |f(x)-L||g(x)| + |L||g(x)-M|.$$ $g$ is bounded near $c$ because the limit $M$ exists. Sending $x\to c$, the right-hand side tends to $0$, as desired. $\square$

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