Asymptote ambiguity

  • Thread starter Didd
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  • #1
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Hello

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

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  • #2
DaveC426913
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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.
 
  • #3
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What does domain restriction have to do with asymptotes ? :confused:

-gaz

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]
 
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  • #4
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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.
 
  • #5
vsage
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.
 
  • #6
28
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Vsage,

I am glad about your responce.

Thank you
 

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