Apostol proof of a basic limit theorem: page 136

Question

In Calculus I by Apostol (page 136), the proof of the statement

$$\lim_{x \to p} \big(f(x)\,g(x)\big) = \big(\lim_{x \to p} f(x)\big)\big(\lim_{x \to p} g(x)\big)$$

is constructed from the special case to the general case. Citing from the book:

Suppose we prove the statement for the special case in which one of the limits is 0. Then the general case follows from this special case. All we need is to write

(A is the limit of $f(x)$, B is the limit of $g(x)$):

$$f(x)g(x) - AB = f(x)(g(x) - B) + B(f(x) - A)$$

The special case implies that each term on the right approaches 0 as $x \rightarrow p$, and the sum of two terms approaches 0. It remains to prove the statement in the special case when one of the limits, say B, is 0.

Can someone help me decipher this?

The problem is the statement about the right side becoming zero. If the limit of $g(x)$ (B) is zero, then the equation $B(f(x) - A)$ indeed becomes zero in the limit. It also becomes zero for ANY limit A of $f(x)$, why does he specify that the either limit has to be zero, when this equation gets 0 with any limits? Moreover, multiplication by the constant in the limit is still a product of functions, so we cannot use it because we are proving it at the moment!

Then, the first part of the rightmost equation, $f(x)(g(x) - B)$, does not make any sense. Even if B is zero, we cannot decide anything about the limit of this compound, since we are proving at the moment the exact multiplication law of limits, and $f(x)(g(x) - B)$ is precisely the limit of a product. So, isn't it a circular logic?

Answer 1

Suppose you somehow manage to prove the following lemma:

Lemma:

Let $\phi$ and $\psi$ be functions from $\Bbb{R}$ into $\Bbb{R}$, such that $\lim_ \limits{x \to p}\phi(x) = 0$ and $\lim_{x \to p} \psi(x) = l$. Then, $\lim \limits_{x \to p} (\phi(x) \cdot \psi(x)) = 0$.

So temporarily, assume that the lemma is true. Now for the general case, we started by assuming that \begin{equation} \lim_{x \to p} f(x) = A \quad \text{and} \quad \lim \limits_{x \to p} g(x) = B \end{equation} This can be equivalently rephrased as \begin{align} \lim_{x \to p} (f(x)- A) = 0 \quad \text{and} \quad \lim \limits_{x \to p} (g(x)- B) = 0. \tag{$*$} \end{align} Now, he performs a simple algebraic manipulation to get \begin{align} f(x) g(x) - AB = f(x) \underbrace{ \left(g(x)- B \right)}_{\phi_1(x)} + B \underbrace{\left( f(x) - A \right)}_{\phi_2(x)} \end{align}

Notice that ($*$) is equivalent to saying $\lim \limits_{x \to p} \phi_1(x) = 0$, and $\lim \limits_{x \to p} \phi_2(x) = 0$. So, we can now apply the lemma to the product, to get: \begin{align} \lim_{x \to p} \left( f(x)g(x) - AB\right) &= \lim_{x \to p}[f(x) \phi_1(x)] + \lim_{x\to p} [B \cdot \phi_2(x)] \tag{sum rule} \\ &= 0 + 0 \tag{by the lemma } \\ &= 0 \end{align} This is exactly what we wanted to prove.

There is nothing circular about this. All he is saying is: let's temporarily assume this lemma, and let me prove a general version of the theorem. After that, we come back and prove the special case.

Answer 2

You can use the function $f(x)(g(x)-B)=f(x)g(x)-f(x)B$, then using the special case it follows that: \begin{equation} \lim_{x\to p} f(x)g(x)-f(x)B=\lim_{x\to p}f(x)(g(x)-B)=\lim_{x\to p} f(x) \lim_{x\to p}(g(x)-B)=A\cdot 0= 0 \end{equation} From the limit of sum you have: \begin{equation} 0=\lim_{x\to p} f(x)g(x)-\lim_{x\to p} f(x)B=\lim_{x\to p} f(x)g(x)-AB\Rightarrow\lim_{x\to p} f(x)g(x)=AB \end{equation} The $\lim_{x\to p} f(x)B=AB$ can justified by: \begin{equation} \lim_{x\to p} f(x)B-AB=\lim_{x\to p} (f(x)-A)B=0 \cdot B=0\Rightarrow \lim_{x\to p} f(x)B=AB. \end{equation}

Answer 3

The argument isn't quite circular. His claim is the following:

"Assume (for the moment) that given any $f$ and $g$ such that $f(x) \to A$ and $g(x) \to 0$ as $x \to p$, we have $f(x)g(x) \to 0$. Then, we know (from this assumption) that, given $h$ and $k$ such that $h(x) \to C$ and $k(x) \to D$ as $x \to p$, we have $h(x)k(x) \to CD$."

In other words, if we assume that we have some proof of the special case of this theorem where $B = 0$, then we will have a proof for the general case, i.e. making no assumptions about $B$. Thus, it just falls to the author to prove the case where $B = 0$, and then by the given argument, the full argument will be completed.

(This is known as making assumptions without loss of generality. The assumption $B = 0$ can be made without losing generality for the full result.)

Answer 4

The logic in Apostol's proof is quite clear. What really confuses you is probably the note under Theorem 3.1 in Apostol's book:

Note: An important special case of (iii) occurs when $f$ is constant, say $f(x)=A$ for all $x$. In this case, (iii) is written as $\lim_{x\to p}A\cdot g(x)=A\cdot B$.

The "special case" in this note is NOT the "special case" in the sentence "Suppose that we have proved part (iii) for the special case in which one of the limits is $0$" that means the following is true:

$$\begin{align} \lim_{x\to p}f_1(x)=0, \lim_{x\to p}g_1(x)=B\ \Longrightarrow \lim_{x\to p}f_1(x)g_1(x)=0\cdot B=0,\tag{1}\\ \lim_{x\to p}f_1(x)=A, \lim_{x\to p}g_1(x)=0\ \Longrightarrow \lim_{x\to p}f_1(x)g_1(x)=A\cdot 0=0.\tag{2} \end{align} $$

Suppose $\lim_{x\to p}f(x)=A$ and $\lim_{x\to p}g(x)=B$ and write $$ f(x)g(x)-AB=f(x)[g(x)-B]+B[f(x)-A]. $$

Then one has by the "special case", i.e., (1) and (2), $$ \lim_{x\to p}f(x)[g(x)-B]=A\cdot 0=0,\quad \lim_{x\to p} B[f(x)-A]=B\cdot 0=0. $$