The problem asks us to determine whether the given statements are true or false. If true, we must explain why. If false, we must provide a counterexample. a) If $a, b \in \mathbb{R} - \{0\}$, then $\ln(ab) = \ln a + \ln b$. b) If $f: A \subseteq \mathbb{R} \to \mathbb{R}$ is a function such that $(c, \infty) \subseteq A$ for some $c \in \mathbb{R}$ and $\lim_{x \to \infty} f(x) = 0$, then $f(x) = 0$ for some $x \in A$.

1. Problem Description

The problem asks us to determine whether the given statements are true or false. If true, we must explain why. If false, we must provide a counterexample.

a) If $a, b \in \mathbb{R} - \{0\}$, then $\ln(ab) = \ln a + \ln b$.

b) If $f: A \subseteq \mathbb{R} \to \mathbb{R}$ is a function such that $(c, \infty) \subseteq A$ for some $c \in \mathbb{R}$ and $\lim_{x \to \infty} f(x) = 0$, then $f(x) = 0$ for some $x \in A$.

2. Solution Steps

a) The statement "If $a, b \in \mathbb{R} - \{0\}$, then $\ln(ab) = \ln a + \ln b$" is false. The identity $\ln(ab) = \ln a + \ln b$ is only true if $a > 0$ and $b > 0$.

If $a$ and $b$ are negative, then $\ln a$ and $\ln b$ are not real numbers.

However, if we consider real numbers, we can examine the case when $a < 0$ and $b < 0$. Then $ab > 0$, so $\ln(ab)$ is defined. But $\ln a$ and $\ln b$ are not defined in the real numbers.

Consider the case where $a=-1$ and $b=-1$. Then $\ln(ab) = \ln((-1)(-1)) = \ln(1) = 0$. But $\ln a = \ln(-1)$ and $\ln b = \ln(-1)$ are undefined in the real numbers.

Another way to look at this problem is to consider the complex logarithm. The complex logarithm satisfies

$\ln(ab) = \ln(a) + \ln(b) + 2\pi i k$, for some integer $k$.

When considering real-valued logarithms, we require $a, b > 0$.

b) The statement "If $f: A \subseteq \mathbb{R} \to \mathbb{R}$ is a function such that $(c, \infty) \subseteq A$ for some $c \in \mathbb{R}$ and $\lim_{x \to \infty} f(x) = 0$, then $f(x) = 0$ for some $x \in A$" is false.

Consider the function $f(x) = \frac{1}{x}$ defined on $A = [1, \infty)$.

Then $(1, \infty) \subseteq A$, and $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{x} = 0$.

However, $f(x) = \frac{1}{x} \neq 0$ for any $x \in [1, \infty)$.

3. Final Answer

a) False. Counterexample: $a=-1$ and $b=-1$. Then $\ln(ab)=\ln(1)=0$, but $\ln(a)$ and $\ln(b)$ are not defined in the real numbers.

b) False. Counterexample: $f(x) = \frac{1}{x}$ defined on $A=[1, \infty)$. Then $\lim_{x \to \infty} f(x) = 0$, but $f(x) \neq 0$ for any $x \in A$.