1. zero divided by zero

Okay, I've checked about ten different sites for an answer to this, and about half said one thing, while the other half disagreed, so I want a definitive answer:

I know that 0/0 is undefined, but what I have are two equations:

f(x)=0

And:

g(x)=0

Where both zeros are of the same density.

Now correct me if I'm wrong, but isn't this a point where I can say:

lim(n approaches x) [(g(n))/(f(n))]=1

2. Originally Posted by rman144
Okay, I've checked about ten different sites for an answer to this, and about half said one thing, while the other half disagreed, so I want a definitive answer:

I know that 0/0 is undefined, but what I have are two equations:

f(x)=0

And:

g(x)=0

Where both zeros are of the same density.

Now correct me if I'm wrong, but isn't this a point where I can say:

lim(n approaches x) [(g(n))/(f(n))]=1
*sigh* Your question is completely vague as posted.

Post the entire question, exactly as it's written in the textbook (or wherever it's from) if you expect anyone to give a relevant reply.

Its pretty self-explanatory:

Does 0/0=1 where both zeros are of the same density?

4. Originally Posted by rman144
Does 0/0=1?
No. And any site that said "yes" or "depends" is an insult to mathematics.

5. Originally Posted by rman144
Its pretty self-explanatory:

Does 0/0=1 where both zeros are of the same density?
TPH is right, the answer is no. the closest thing to what you may want is a limit existing if both the numerator and denominator go to zero, for example $\displaystyle \lim_{x \to 0} \frac {\sin x}x = 1$. but of course, this is not always true, $\displaystyle \lim_{x \to 0} \frac {1 - \cos x}x = 0$. so limits might exists, but will not always be 1, they can be anything, however 0/0 is always undefined

6. rman144, maybe you are confused about the concept of limit of the quotient of 2 functions that tends to 0 in a point and the number 0 divided by 0. What you call "same density" would be that the 2 functions tends to 0 at almost the same "speed".
For example, if $\displaystyle f(x)=x$ and $\displaystyle g(x)= \sin(x)$. The limit of the quotient of both function will give $\displaystyle 1$ when $\displaystyle x$ tends to $\displaystyle 0$. But it doesn't mean that because $\displaystyle \sin (0)=0$ and $\displaystyle 0=0$ the limit of the quotient is equal to $\displaystyle 0$ over $\displaystyle 0$. If you understand the concept of limit, you know it's not true so you cannot get confused on this.
EDIT:Jhevon, you are fast! I didn't know someone would answer that fast, sorry for my post.
Back to the concept of limit : rman144. In my example, say we have $\displaystyle \frac{\sin (x)}{x}$. As x approaches 0, the quotient approaches 1. You can get closer and closer to 1 when you chose an x closer and closer to 0. But it doesn't imply that if you chose 0 as x that you'll get 1 as the quotient! Be careful on this.

So as a follow-up question, what does the following equal:

Given two equations:

f(x) and g(x)

Where:

f(n)=g(n)=0

Find:

(f(n))/(g(n))

This would be unfined, correct?

8. So as a follow-up question, what does the following equal:

Given two equations:

f(x) and g(x)

Where:

f(n)=g(n)=0

Find:

(f(n))/(g(n))

This would be unfined, correct?
Yes! Because it will be equal to 0 over 0 which is undefined.

9. Originally Posted by mr fantastic
*sigh* Your question is completely vague as posted.

Post the entire question, exactly as it's written in the textbook (or wherever it's from) if you expect anyone to give a relevant reply.
I repeat. Perhaps this time there will be an appropriate response .....

10. I think this discussion has covered the OP's question. As for anything else, this topic has been discussed ad nauseum elsewhere on this site.

-Dan