lim f(x)*lim g(x)=lim[f(x)*g(x)]

**james_bond** · November 11th 2008, 08:53 AM

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.

**HallsofIvy** · November 11th 2008, 10:32 AM

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|< $\epsilon/2+ \epsilon/2+ \epsilon$

**james_bond** · November 12th 2008, 05:05 AM

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.

**CaptainBlack** · November 12th 2008, 05:25 AM

Originally Posted by james_bond

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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