Proof: limit of product is the product of limits

[nuuskur]

Homework Statement

Let $f_1,f_2\colon\mathbb{R}^m\to\mathbb{R}$ and a cluster point $P_0\in D\subset\mathbb{R}^m$ (domain)
Prove that $\lim_{P\to P_0} f_1(P)\cdot f_2(P) = \lim_{P\to P_0} f_1(P)\cdot\lim_{P\to P_0} f_2(P)$

Homework Equations

The Attempt at a Solution

Let $\begin{cases} \lim_{P\to P_0} f_1(P) = A \\ \lim_{P\to P_0} f_2(P) = B\end{cases}$
As $P_0$ is a cluster point, there exists a sequence $(P_n)$ such that $\lim_n P_n = P_0$

Is this correct? A cluster point in the domain is a point whose every ball around it intersects with the domain, hence the sequence should exist.
For every $\varepsilon > 0$ there exists $B(P_0,\varepsilon)$ such that $(B(P_0,\varepsilon)\setminus \{P_0\})\cap D\neq\emptyset$

Per the sequential criterion for limits the 2 statements are equivalent:
1) $\lim_{P\to P_0} f(P) = L$
2) If $\left [P_n\in D\setminus \{P_0\}, n\in\mathbb{N}\colon \lim_{n} P_n = P_0 \right ]$ then $\lim_{n} f(P_n) = L$
I am curious why I was suggested to Not use $\forall, \exists, \Rightarrow$ and such if they were made for that exact purpose.

We know the sequence $(P_n)$ exists such that $P_n\xrightarrow[n\to\infty]{}P_0$. It should suffice to show that $f_1(P_n)f_2(P_n)\xrightarrow[n\to\infty]{} AB$. (?)

Assume 1) is valid and let $P_n\in D\setminus \{P_0\}$ such that $P_n\xrightarrow[n\to\infty]{}P_0$. Let $\varepsilon > 0$ then there exists an index $N\in\mathbb{N}$ such that
$$n\geq N\Rightarrow |f_1(P_n)f_2(P_n) - AB| < \varepsilon$$
As $f_1(P)f_2(P)\xrightarrow[P\to P_0]{} AB$ then there exists $\delta > 0$ such that $$0 < d(P,P_0) < \delta \Rightarrow |f_1(P)f_2(P) - AB| < \varepsilon$$
Knowing that $P_n\xrightarrow[n\to\infty]{}P_0$ then there exists and index $N\in\mathbb{N}$ such that
$$n\geq N \Rightarrow d(P_n, P_0)< \delta$$
Therefore: if $n\geq N$ then $d(P_n,P_0) < \delta$ and $|f_1(P_n)f_2(P_n) - AB|<\varepsilon$

Assume 2) is valid and assume by contradiction that 1) is not valid then there exists $\varepsilon > 0$ such that for every index $n\in\mathbb{N}$ there exists a point $P_n\in D\setminus \{P_0\}$ such that $$d(P_n, P_0) < \frac{1}{n},\ \mathrm{but}\ \ \ |f_1(P_n)f_2(P_n) - AB| \geq\varepsilon$$
However, $P_n\xrightarrow[n\to\infty]{}P_0$ and not $f_1(P_n)f_2(P_n)\xrightarrow[n\to\infty]{}AB$ contradicts the validity of 2) $_{\blacksquare}$

Is this enough to show that the limit of a product is the product of limits?

4. Additional notes

How can I show this the usual way, without the sequence criterion?

Let $f_1(P)\xrightarrow[P\to P_0]{} A$ and $f_2(P)\xrightarrow[P\to P_0]{} B$

$f_1\colon \forall\varepsilon > 0,\exists\delta_1 > 0\ \ |\ \ 0 < d(P,P_0) < \delta_1 \Rightarrow |f_1(P) - A| < \varepsilon$

$f_2\colon \forall\varepsilon > 0,\exists\delta_2 > 0\ \ |\ \ 0 < d(P,P_0) < \delta_2 \Rightarrow |f_2(P) - B| < \varepsilon$

$f_1\cdot f_2\colon \forall\varepsilon > 0,\exists\delta > 0\ \ |\ \ 0 < d(P,P_0) < \delta \Rightarrow |f_1(P)f_2(P) - AB| < \varepsilon$

Essentially I have to show that:
$|f_1(P) - A| |f_2(P) - B|<\varepsilon$ is somehow equivalent to $|f_1(P)f_2(P) - AB|<\varepsilon$
I have $|(f_1(P) - A)(f_2(P) - B)| < \varepsilon$. How do I choose the epsilon so it would give me the desired result?

 

[micromass]

Homework Statement

Let $f_1,f_2\colon\mathbb{R}^m\to\mathbb{R}$ and a cluster point $P_0\in D\subset\mathbb{R}^m$ (domain)
Prove that $\lim_{P\to P_0} f_1(P)\cdot f_2(P) = \lim_{P\to P_0} f_1(P)\cdot\lim_{P\to P_0} f_2(P)$

Can you tell me why $P_0$ being a cluster point is important? It is not really important for the proof (see later).

Let $\begin{cases} \lim_{P\to P_0} f_1(P) = A \\ \lim_{P\to P_0} f_2(P) = B\end{cases}$
As $P_0$ is a cluster point, there exists a sequence $(P_n)$ such that $\lim_n P_n = P_0$

Is this correct? A cluster point in the domain is a point whose every ball around it intersects with the domain, hence the sequence should exist.
For every $\varepsilon > 0$ there exists $B(P_0,\varepsilon)$ such that $(B(P_0,\varepsilon)\setminus \{P_0\})\cap D\neq\emptyset$

Yes, this is correct. But what follows is not correct. The issue is that you picked a particular sequence $(P_n)_n$, while the argument later should hold for any sequence $(P_n)_n$. So you should remove the sentence "As $P_0$ is a cluster point, there exists a sequence $(P_n)$ such that $\lim_n P_n = P_0$" from your proof.

Per the sequential criterion for limits the 2 statements are equivalent:
1) $\lim_{P\to P_0} f(P) = L$
2) If $\left [P_n\in D\setminus \{P_0\}, n\in\mathbb{N}\colon \lim_{n} P_n = P_0 \right ]$ then $\lim_{n} f(P_n) = L$
I am curious why I was suggested to Not use $\forall, \exists, \Rightarrow$ and such if they were made for that exact purpose.

It is to make your proofs more readable. It really makes a huge difference to people reading your proof! I would even rewrite your sentence as "For any sequence $(P_n)_n$ in $D\setminus \{P_0\}$ that converges to $P_0$, we have that $\lim_n f(P_n) = L$.

We know the sequence $(P_n)$ exists such that $P_n\xrightarrow[n\to\infty]{}P_0$. It should suffice to show that $f_1(P_n)f_2(P_n)\xrightarrow[n\to\infty]{} AB$. (?)

No, the existence of the sequence is irrelevant here. You need to prove it for any sequence, not a particular one. So in the sequel I will assume that $(P_n)_n$ is an arbitrary sequence in $D\setminus \{P_0\}$ that converges to $P_0$ (and thus not necessarily the one that exists from the limit point thing).

Assume 1) is valid

Why are you showing the equivalence of (1) and (2)? I assume you know this already?

and let $P_n\in D\setminus \{P_0\}$ such that $P_n\xrightarrow[n\to\infty]{}P_0$. Let $\varepsilon > 0$ then there exists an index $N\in\mathbb{N}$ such that
$$n\geq N\Rightarrow |f_1(P_n)f_2(P_n) - AB| < \varepsilon$$

Please don't use $\Rightarrow$

As $f_1(P)f_2(P)\xrightarrow[P\to P_0]{} AB$

This is what you need to prove. You can't state it and use it.

How can I show this the usual way, without the sequence criterion?

Let $f_1(P)\xrightarrow[P\to P_0]{} A$ and $f_2(P)\xrightarrow[P\to P_0]{} B$

$f_1\colon \forall\varepsilon > 0,\exists\delta_1 > 0\ \ |\ \ 0 < d(P,P_0) < \delta_1 \Rightarrow |f_1(P) - A| < \varepsilon$

Whatever you do, certainly NEVER use $|$ in this context. It is only valid in set-builder notation like $\{x~|~x\notin x\}$. It is never used outside it.

Anyway, a hint or the proof:

$|f_1(P)f_2(P) - AB| = |f_1(P)f_2(P) -Af_2(P) + Af_2(P) - AB| \leq |f_1(P) - A| |f_2(P)| + A|f_2(P) - B|$