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:

[1] https://tutorial.math.lamar.edu/classes/calci/limitproofs.aspx

About Author

I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.

You might also enjoy: