2

The problem is as follows:

Let c be a cluster point of A subset R and suppose that f: A -> R and g: A -> R are functions such that the limits of f(x) and g(x) exist as x goes to c. Prove that:

$$lim(f(x)g(x)) = (limf(x))(lim(g(x))$$

as x goes to c.

I just solved a problem which is apparently easier than this one:

Prove that:

$$ lim(f(x)+g(x)) = lim(f(x)) + lim(g(x)$$

as x goes to c

proof:

Take sequence {$x_n$} -> c with $x_n \ne$ c for all n then:

{f($x_n$)} -> lim(f(x)) = $L_1$

{g($x_n$)} -> lim(g(x)) = $L_2$

then lim(f($x_n$) + g($x_n$)) = lim f($x_n$) + lim g($x_n$) = lim f(x) + lim g(x)

since {(f+g)($x_n$)} -> $L_1$ + $L_2$ for every {$x_n$} -> c

then the original statement holds by the sequential criterion.

I was wondering if the same logic can be applied to the original problem I asked. So if I were to replace the + signs with * would it be prove the first problem?

Community
  • 1
Adagio
  • 379
  • 1
  • 5

1 Answers1

2

It is sufficient to consider the sequence case:

Consider two sequences, $\left\{a_n\right\}_{n=1}^{\infty}$ and $\left\{b_n\right\}_{n=1}^{\infty}$, such that:

$$\lim_{n\to\infty}a_n=L \ and \ \lim_{n\to\infty}b_n=M$$

Then $\forall \epsilon >0, \exists \ N_1,N_2\in \mathbb{N}$ such that:

$$n>N_1\Rightarrow |a_n-L|<\frac {\epsilon}{(B+|L|)}=\epsilon_1, \ and \ n>N_2\Rightarrow|b_n-M|<\frac {\epsilon}{(B+|L|)}=\epsilon_1$$

Take $n>max\left\{N_1,N_2\right\}$

So consider the following:

$$|a_nb_n-LM|=|a_nb_n-Lb_n+Lb_n-LM|=|b_n||a_n-L|+|L||b_n-M|$$

Because $\left\{b_n\right\}_{n=1}^{\infty}$ is convergent with limit $M$, it is bounded. Let's call that bound $B$.

So,

$$|a_nb_n-LM|\leq|B||a_n-L|+|L||b_n-M|=\epsilon\frac {(|B|+|L|)}{(|B|+|L|)}=\epsilon$$

Thus, $\lim_{n\to\infty}(a_n\cdot b_n)=LM \ \exists$

Mark Pineau
  • 1,400