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Math Help - lim f(x)*lim g(x)=lim[f(x)*g(x)]

  1. #1
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    lim f(x)*lim g(x)=lim[f(x)*g(x)]

    Prove that \lim_{x\rightarrow\infty}\left[f(x)\cdot g(x)\right]=\lim_{x\rightarrow\infty}f(x)\cdot\lim_{x\rightar  row\infty}g(x) using only the definition.
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  2. #2
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    Let F= \lim_{x\rightarrow \infty} f(x) and G= \lim_{x\rightarrow \infty} g(x)

    Now look at |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ FG| \le|f(x)g(x)- Fg(x)|+ |Fg(x)- FG|.

    |f(x)g(x)- F(g)|= |g(x)||f(x)-F| and |Fg(x)- FG|= |F||g(x)- G|.

    Since G= \lim_{x\rightarrow \infty}, for any \nu> 0 there exist N_1 such that if x> N_1, |g(x)- G|< \nu. In particular, given \epsilon> 0, we can take \nu= 2\epsilon/|F| so that |g(x)- G|< \epsilon/|F| and |F||g(x)- G|< \epsilon/2.

    g(x)|f(x)- F| is a little harder because g(x) is a variable. However, since g(x) converges to G, for x greater than some N_2, |g(x)-G|< 1 and so G-1< g(x)< G+1. In either case |g(x)|< M where M is the larger of |G-1| and |G+1|. In that case |g(x)||f(x)- F|< M|f(x)- F| and, just as before, given any \epsilon> 0, there exist N_3 such that if x> N_3, |M||f(x)- F|< \epsilon/2[/tex].

    Putting those together, as long as x> the largest of N_1, N_2, and N_3, all those are true and |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ FG| \le|f(x)g(x)- Fg(x)|+ |Fg(x)- FG|< \epsilon/2+ \epsilon/2+ \epsilon
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  3. #3
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    My notes:

    |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ Fg(x)- FG| \le|f(x)g(x)- Fg(x)|+ |Fg(x)- FG|

    |f(x)g(x)- Fg(x)|= |g(x)||f(x)-F|

    G= \lim_{x\rightarrow \infty}\color{red}g(x) ... we can take \nu= \epsilon/\color{red}(2\color{black}|F|\color{red})\color{bl  ack} so that |g(x)- G|< \epsilon/\color{red}(2\color{black}|F|\color{red})\color{bl  ack}

    ... given any \epsilon \color{red}/2\color{black}> 0, there exist N_3 such that if x> N_3, |M||f(x)- F|<  \epsilon/2.

    ... |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ Fg(x)- FG| \le|f(x)g(x)- Fg(x)|+ |Fg(x)- FG|< \epsilon/2+ \epsilon/2\color{red}=\color{black} \epsilon

    Why don't you write the whole equations in LaTeX why just only the critical parts of them? Otherwise thanks for the proof.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by james_bond View Post
    Prove that \lim_{x\rightarrow\infty}\left[f(x)\cdot g(x)\right]=\lim_{x\rightarrow\infty}f(x)\cdot\lim_{x\rightar  row\infty}g(x) using only the definition.
    You need the assumption that both the limits on the right exist.

    CB
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