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

    Now look at |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ FG|$\displaystyle \le$|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 $\displaystyle G= \lim_{x\rightarrow \infty}$, for any $\displaystyle \nu> 0$ there exist $\displaystyle N_1$ such that if $\displaystyle x> N_1$, |g(x)- G|< $\displaystyle \nu$. In particular, given $\displaystyle \epsilon> 0$, we can take $\displaystyle \nu= 2\epsilon/|F|$ so that |g(x)- G|< $\displaystyle \epsilon/|F|$ and |F||g(x)- G|< $\displaystyle \epsilon/2$.

    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 $\displaystyle N_2$, |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 $\displaystyle \epsilon> 0$, there exist $\displaystyle N_3$ such that if $\displaystyle x> N_3$, |M||f(x)- F|< \epsilon/2[/tex].

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

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

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

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

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

    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 $\displaystyle \lim_{x\rightarrow\infty}\left[f(x)\cdot g(x)\right]=\lim_{x\rightarrow\infty}f(x)\cdot\lim_{x\rightar row\infty}g(x)$ using only the definition.
    You need the assumption that both the limits on the right exist.

    CB
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