In this post, I show that $\lim_{x\rightarrow a}[f(x)g(x)] = \lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$ given that $\lim_{x\rightarrow a} f(x) = A$ and $\lim_{x\rightarrow a} g(x) = B$. To do this, I approximately follow the steps in reference [1]. Known:  Using the definition of a limit, $|f(x) – A|<\epsilon_1$ whenever $ 0 < |x-a| < \delta$, with $\epsilon_1 > 0$. $|g(x)- B|<\epsilon_2$ whenever $ 0 < |x-a| < \delta$, with $\epsilon_2 > 0$. Objective: The objective is to directly show that $\lim_{x\rightarrow a}[f(x)g(x)] = \lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$. Proof: Using algebra, $ [f(x) – A] [g(x) – B] = f(x)g(x) – f(x)B – Ag(x) + AB$ Solving for $f(x)g(x)$, $ [f(x) – A] [g(x) – B] + f(x)B + Ag(x) – AB = f(x)g(x)$ Substituting, $ \lim_{x\rightarrow a}\big(f(x)g(x)\big) = \lim_{x\rightarrow a} \big( [f(x) – A] [g(x) – B] + f(x)B + Ag(x) – AB \big)$ Using this property and this property, $ = \lim_{x \rightarrow a} [f(x) – A] [g(x) – B] + \lim_{x\rightarrow a}f(x)B $ $ + \lim_{x\rightarrow a}Ag(x) – \lim_{x\rightarrow a}AB $ From this property, $ \lim_{x\rightarrow a}AB = AB$ And from this property, $ \lim_{x\rightarrow a}f(x)B = B\lim_{x\rightarrow a}f(x) = AB$ $ \lim_{x\rightarrow a}Ag(x) = A\lim_{x\rightarrow a}g(x)$ Substituting these three results, $ \lim_{x\rightarrow a}\big(f(x)g(x)\big) = \lim_{x \rightarrow a} [f(x) – A] [g(x) – B] + A\lim_{x\rightarrow a}g(x)$ Examine two limits using the above properties: $\lim_{x \rightarrow a} [f(x) – A] = \lim_{x \rightarrow a} f(x) – \lim_{x \rightarrow a} A = A – A = 0$ $\lim_{x \rightarrow a} [g(x) – B] = \lim_{x \rightarrow a} g(x) – \lim_{x \rightarrow a} B = B – B = 0$ Introduce a new expression and use this absolute-value property: $| \big(  f(x) – A   \big) \big(  g(x) – B   \big)| = | f(x) – A  | |  g(x) – B  | $ The left side of the previous line is equal to $| \big(  f(x) […]