# Proving division of continuous functions

1. Nov 26, 2007

### buZZ

1. The problem statement, all variables and given/known data

Okay, so if f and g are continuous functions at a, then prove that f/g is continuous at a if and only if g(a) # 0

2. Relevant equations
Assuming to start off the g(a)#0, by the delta-epsilon definition of continuity, basically, We know that |f(x)| and |g(x)| are bounded.

3. The attempt at a solution

I have messed around with the end result that we need, which is when |x-a|< delta
|f(x)/g(x)-f(a)g(a)|<Epsilon. This is what I've come up with:

|1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|.
By looking at each piece it seems like they can be bounded as well. However, how do i manipulate what each one is bounded by so that when I multiply and add everything out, I get a nice simple Epsilon?

2. Nov 26, 2007

### buZZ

you know...I could be going about it completely the wrong way too..

3. Nov 26, 2007

### Hurkyl

Staff Emeritus
You've already proven a bunch of limit theorems. Can you use those limit theorems to derive this one?

P.S. I assume by # you meant $\neq$; you should say that someplace rather than just using it without warning.

4. Nov 26, 2007

### Hurkyl

Staff Emeritus
If both pieces are bounded by $\sqrt{\epsilon}$, then what is their product bounded by?

(I assume by "piece" you meant the two multiplicands)

5. Nov 26, 2007

### buZZ

sorry about that, I didn't realize you could do that...I'm new here,
And I need to use the rigorous definitions of continuity, so deriving it here from what I know about the bunches of limit theorems I know isn't sufficient.

6. Nov 26, 2007

### buZZ

It doesn't seem that easy. Maybe I should back up.
The fact that f and g are continuous at a mean that for |x-a|<delta,
|f(x)-f(a)|<Something involving epsilon(1)
|g(x)-g(a)|<something involving ep(2)

By triangle inequality,
|f(x)|< something involving E(1) + |f(a)|
|g(x)|< something involving E(2) + |g(a)|

Since this is what we know, we have to start multiplying around and around using these two inequalities to get the final result |1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|. For example,
|f(x)g(a)|< something involving E(1)*|g(a)|+|f(a)g(a)|.

Now I'm confusing myself again, Am I making this too complicated? or what

7. Nov 26, 2007

### Hurkyl

Staff Emeritus
There's a famous trick that's very useful for manipulating absolute value inequalities:

AB - CD = AB - AD + AD - CD = A(B-D) + (A-C)D

So if B-D and A-C are things you happen to know something about...

8. Nov 26, 2007

### buZZ

that only seems useful if I'm trying to prove that f*g is continuous at a, but I'm lost as to see why that's useful for f/g

9. Nov 27, 2007

### Office_Shredder

Staff Emeritus
Because you have

|1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|

Specifically, the
|f(x)g(a)-f(a)g(x)|

term can be split. to deal with the 1/g(x), note that eventually g(x)>g(a)-e (e being epsilon) for all x within delta of a, for any epsilon

10. Nov 28, 2007

### buZZ

Okay, abandon my initial approach. I think it'd be ten times easier if I show that 1/g(x) is continuous at a. Since g(a) is not zero, then g(x) has to bounded away from zero. Which means that 1/g(x) will not get infinitely large, but rather, will be bounded by something else. Now if only I could put that into a nice algebraic argument that would have a bit more clout.
If |g(x)-g(a)|< e then how do you get that g(x) > g(a) - e without ignoring the absolute values?

11. Nov 28, 2007

### Office_Shredder

Staff Emeritus
By definition, that means -e< g(x)-g(a) < e so g(a)-e<g(x) and g(x)< g(a)+e

Geometric intuition time: If |x-a|<e for any kind of x, a and e, that means that x is within e of a.