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Homework Help: Proof: limit of product is the product of limits

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f_1,f_2\colon\mathbb{R}^m\to\mathbb{R} [/itex] and a cluster point [itex]P_0\in D\subset\mathbb{R}^m[/itex] (domain)
    Prove that [itex]\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) [/itex]

    2. Relevant equations

    3. The attempt at a solution
    Let [itex]\begin{cases} \lim_{P\to P_0} f_1(P) = A \\ \lim_{P\to P_0} f_2(P) = B\end{cases}[/itex]
    As [itex]P_0[/itex] is a cluster point, there exists a sequence [itex](P_n)[/itex] such that [itex]\lim_n P_n = P_0[/itex]

    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 [itex]\varepsilon > 0[/itex] there exists [itex]B(P_0,\varepsilon)[/itex] such that [itex](B(P_0,\varepsilon)\setminus \{P_0\})\cap D\neq\emptyset[/itex]

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

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

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

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

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

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

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

    [itex]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[/itex]

    Essentially I have to show that:
    [itex]|f_1(P) - A| |f_2(P) - B|<\varepsilon[/itex] is somehow equivalent to [itex]|f_1(P)f_2(P) - AB|<\varepsilon [/itex]
    I have [itex] |(f_1(P) - A)(f_2(P) - B)| < \varepsilon [/itex]. How do I choose the epsilon so it would give me the desired result?
    Last edited: Apr 3, 2015
  2. jcsd
  3. Apr 4, 2015 #2
    Can you tell me why ##P_0## being a cluster point is important? It is not really important for the proof (see later).

    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 [itex]P_0[/itex] is a cluster point, there exists a sequence [itex](P_n)[/itex] such that [itex]\lim_n P_n = P_0[/itex]" from your proof.

    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##.

    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).

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

    Please don't use ##\Rightarrow##

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

    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|##
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