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:

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