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

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*}

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. $$\square$$

• Exactly how do you go about simplifying it using "algebraic limit laws"? Jun 13 '21 at 0:23
• @Peatherfed: I added the laws I was thinking about. Once you have these in mind, to finish the proof of the general case using the special case when the limits are zero, use the usual algebraic laws like distributivity, commutativity, and so on, and then apply the two bulleted laws when applicable to get the final result. Hope this helps. Jun 13 '21 at 1:20
• I see, I was hoping to actually get that second law as a consequence of the theorem. I might go about it by first showing it for constants, which will be almost immediate, and then proceeding through the special case into the general case using the linearity of limits. As concerns your suggestion for the direct proof, I like it, although I was aiming for the type of proof where the delta is explicitly provided. I essentially use that g is bounded when going from the first to the second inequality. Jun 13 '21 at 1:31

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.

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.

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.