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

Question

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*}

Answer 1

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. $\square$

Answer 2

It may be of interest to you that (with some abstract tools from model theory) one can choose to use nonstandard analysis instead of $\epsilon-\delta$ and work with infinitesimal numbers. I could imagine that the book "Nonstandard Analysis" by Martin Väth contains basic calculus rules proven within the framework of nonstandard analysis.

Answer 3

Note that you don't really need to split in two cases, you can do it all at once: It's easy to see that $g$ is bounded near $x=c$ (because the limit exists), say $|g(x)|\leq K$, $K>0$, for $|x-c|<\delta_{3}$. Given $\varepsilon>0$ there exists $\delta_{1},\delta_{2}>0$ such that $$|x-c|<\delta_{1}\Rightarrow |f(x)-L|<\frac{\varepsilon}{2K}\text{ and }|x-c|<\delta_{2}\Rightarrow |g(x)-M|<\frac{\varepsilon}{2(|L|+1)}.$$ Now take $\delta=\min\{\delta_{1},\delta_{2},\delta_{3}\}$. For $|x-c|<\delta$ we have that \begin{align*} |f(x)g(x)-LM|&\leq |f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &\leq |g(x)|\cdot |f(x)-L|+|L|\cdot |g(x)-M|\\ &< K\cdot\frac{\varepsilon}{2K}+\frac{|L|\cdot\varepsilon}{2(|L|+1)}\\ &<\varepsilon \end{align*} and we are done.

Answer 4

Some time ago I found a very clean proof in Landau's Differential Calculus book which showed the product of continuous functions is continous.

The style was a bit outdated but I managed to update it and the idea works just fine with limits.

The advantage of the approach is that you don't have to show the function is bounded in a neighbourhood of the point. Furthermore you don't have to break into cases where $L$ and $M$ are zero or not.

First notice that

$$\begin{array}{c} |(fg)(x)-LM|&=|f(x)g(x)-f(x)M-Lg(x)+LM+Lg(x)-LM+Mf(x)-ML|\\ &=|(f(x)-L)(g(x)-M)+L(g(x)-M)+M(f(x)-L)|\\ &\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L| \end{array}$$ for every $x$. Now, given $\varepsilon>0$ there is $\delta_1>0$ such that

$$0<|x-a|<\delta_1\implies |f(x)-L|<\min\left\{1, \frac{\varepsilon}{3(1+|M|)}\right\}$$

and there is $\delta_2>0$ such that:

$$0<|x-a|<\delta_2\implies |g(x)-M|<\frac{\varepsilon}{3(1+|L|)}.$$ Now notice,

$$\begin{array}{c} 1<1+|L|&\implies \frac{1}{1+|L|}<1\\ |L|<1+|L|&\implies \frac{|L|}{1+|L|}<1\\ |M|<1+|M|&\implies \frac{|M|}{1+|M|}<1. \end{array}$$ Using this, it follows that

$$0<|x-a|<\min\{\delta_1, \delta_2\}$$ implies $$ \begin{align*} |(fg)(x)-LM|&\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L|\\ &<1\cdot \frac{\varepsilon}{3(1+|L|)}+|L|\frac{\varepsilon}{3(1+|L|)}+|M|\frac{\varepsilon}{3(1+|M|)}\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon. \end{align*}$$

The same idea also applies when you're working with sequences, for instance.