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

$| \big( f(x) – A \big) \big( g(x) – B \big) – 0|$

so

$| \big( f(x) – A \big) \big( g(x) – B \big) – 0| = | f(x) – A | | g(x) – B |$

Next, examine the right side of the previous line. Since

1. $| f(x) – A | < \epsilon_1$
2. $| g(x) – B | < \epsilon_2$
3. $\epsilon_1 > 0$
4. $\epsilon_2 > 0$
5. A positive number multiplied by another positive number is a positive number,

the following inequalities are satisfied:

$| f(x) – A | | g(x) – B | < \epsilon_3$ with $\epsilon_3 > 0$.

From the transitive property,

$| \big( f(x) – A \big) \big( g(x) – B \big) – 0| < \epsilon_3$ with $\epsilon_3 > 0$

whenever $0 < |x-a| < \delta$.

More compactly, $\lim_{x \rightarrow a} \big( f(x) – A \big) \big( g(x) – B \big) = 0$.

Recall that

$\lim_{x\rightarrow a}\big(f(x)g(x)\big) = \lim_{x \rightarrow a} [f(x) – A] [g(x) – B] + A\lim_{x\rightarrow a}g(x)$.

Using $\lim_{x \rightarrow a} [ f(x) – A ] [g(x) – B ] = 0$ and $\lim_{x \rightarrow a} f(x) = A$,

$\lim_{x\rightarrow a} \big(f(x)g(x)\big) = A\lim_{x\rightarrow a}g(x) = \lim_{x \rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$.

Therefore,

$\lim_{x\rightarrow a}f(x)g(x) = \lim_{x \rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$.

$\square$

References: