Prove that using only the definition.

Results 1 to 4 of 4

- Nov 11th 2008, 08:53 AM #1

- Joined
- Nov 2007
- Posts
- 329

- Nov 11th 2008, 10:32 AM #2

- Joined
- Apr 2005
- Posts
- 19,243
- Thanks
- 2835

Let and

Now look at |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ FG| |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 , for any there exist such that if , |g(x)- G|< . In particular, given , we can take so that |g(x)- G|< and |F||g(x)- G|< .

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 , |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 , there exist such that if , |M||f(x)- F|< \epsilon/2[/tex].

Putting those together, as long as x> the largest of , , and , all those are true and |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ FG| |f(x)g(x)- Fg(x)|+ |Fg(x)- FG|<

- Nov 12th 2008, 05:05 AM #3

- Joined
- Nov 2007
- Posts
- 329

My notes:

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

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

... we can take so that |g(x)- G|<

... given any , there exist such that if , |M||f(x)- F|< .

... |f(x)g(x)- FG|= |f(x)g(x)- Fg(x)+ Fg(x)- FG| |f(x)g(x)- Fg(x)|+ |Fg(x)- FG|<

Why don't you write the whole equations in LaTeX why just only the critical parts of them? Otherwise thanks for the proof.

- Nov 12th 2008, 05:25 AM #4

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5