1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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:

    http://abyss.zapto.org/temp/Asymptote.pdf [Broken]
    Last edited by a moderator: May 1, 2017
  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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook