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Asymptote ambiguity

  1. Nov 17, 2004 #1

    Look for the attachment and if you found any ambiguity, please reply.

    Attached Files:

    Last edited: Nov 17, 2004
  2. jcsd
  3. Nov 17, 2004 #2


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    Gold Member

    You're crazy (or malicious) if you think I'm going to open the Word doc of a complete stranger.

    Don't be lazy - post it.
  4. Nov 18, 2004 #3
    What does domain restriction have to do with asymptotes ? :confused:


    btw i exported to pdf because i have no life and should be studying. :P aah the wonders of procrastination. It's amazing how clean your room gets when you have exams :rolleyes:

  5. Nov 20, 2004 #4
    It is informal defination. Truly, it confuses a lot if you are not patient enough to grasp the idea.

    At this time, I was unable to delete the message. On other time, I might resend it with clear and formal way of prooving it. What I prsented there is , "Informal way of prooving". I appolgise for that. I on my way, was trying to make my idea very easy but the outcome is unattractive.
  6. Nov 20, 2004 #5
    What you wrote looks OK to me but it's definitely not what I'm used to seeing in a proof (you're right it was very "informal") and I have a few holes to fill in (pun intended). You say that asymptotes are *the* restriction of the domain of the domain/range of a function. You also say that if the numerator of f(x) = 0 then f(x) has no asymptotes and, by your definition, no restrictions in the domain. What if the denominator is a polynomial of x of degree greater than 0? For example, f(x) = 0/(x-1) is not defined at x = 1 and therefore has a domain restriction (but you're right it isn't an asymptote) so I think you may want to refine your alternate definition of an asymptote a little. I like most of the rest of it though.
  7. Nov 21, 2004 #6

    I am glad about your responce.

    Thank you
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