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Math Help - zero divided by zero

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by rman144 View Post
    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.
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  3. #3
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    Reply

    Its pretty self-explanatory:

    Does 0/0=1 where both zeros are of the same density?
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    Quote Originally Posted by rman144 View Post
    Does 0/0=1?
    No. And any site that said "yes" or "depends" is an insult to mathematics.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rman144 View Post
    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 \lim_{x \to 0} \frac {\sin x}x = 1. but of course, this is not always true, \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
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  6. #6
    MHF Contributor arbolis's Avatar
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    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 f(x)=x and g(x)= \sin(x). The limit of the quotient of both function will give 1 when x tends to 0. But it doesn't mean that because \sin (0)=0 and 0=0 the limit of the quotient is equal to 0 over 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 \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.
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    Reply

    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?
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  8. #8
    MHF Contributor arbolis's Avatar
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    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.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    *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 .....
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  10. #10
    Forum Admin topsquark's Avatar
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    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
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