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-   -   Topic: infinite limits (https://www.askmehelpdesk.com/showthread.php?t=374991)

  • Jul 13, 2009, 05:58 AM
    thinay
    Topic: infinite limits
    Can anyone help me with this problem. :)

    The given is: g(x) = x/(x-5) and I need to find the vertical asymptote of the graph of g and the sketch of the graph of g for values close to the asymptote.

    Here is how I solved it.
    g(x) = x/(x-5)
    x->5
    = (lim x) / (lim x - lim 5)
    x->5 x->5
    = 5/(5-5)
    = 5/0
    = positive infinity

    I don't know if my answer is correct but I tried my best to answer it. I don't have a graph yet. :(

    I hope you can help me with this. Thanks in advance! :)
  • Jul 13, 2009, 06:14 AM
    Unknown008

    You have to find at which value of x does g(x) equals to positive infinity. That value is the line where you have the vertical asymptote.

    The graphing should be quite easy.
  • Jul 13, 2009, 06:25 AM
    thinay

    Wait. I want to know first if my answer 5/0 = positive infinity correct? :)
  • Jul 13, 2009, 06:28 AM
    Unknown008

    Yup, so what you think the value of x is? (you already got that! ;))
  • Jul 13, 2009, 06:32 AM
    thinay

    Is x=5?? :D
  • Jul 13, 2009, 06:34 AM
    Unknown008

    YES!! Congrats! :)
  • Jul 13, 2009, 06:43 AM
    thinay

    Thank you for your help! Clap! Clap! Clap! :D

    wait. To get the values close to the asymptote.. Am I just going to substitute the value of x..

    example: g(x) = (5.01)/(5.01-5)
    is this correct? :) or not? :(
  • Jul 13, 2009, 06:52 AM
    Unknown008

    You can do that. However, if you're plotting, that might be somewhat difficult.

    If you did the , that gives 501. Some larger numbers may perhaps be more practical. Still, I don't know how you do in your country to do such numbers, but you'll see that the graph decreases very fast, then it becomes more horizontal around x=8
  • Jul 13, 2009, 06:59 AM
    thinay

    Oh! I see. Thanks a lot! :D
  • Jul 13, 2009, 07:03 AM
    Unknown008

    And you can try plotting the part to the right of the asymptote. You'll get something that you have perhaps never seen when you go to the negative values of x.
  • Jul 13, 2009, 07:14 AM
    thinay

    Ok! Ok! I'll do that. :)

    Hey! :D What do you think in this problem? Is it positive or negative infinity?
    lim (x/x+2) + (1/x^2-4)
    x->-2^+

    when I tried to answer it. I end with + infinity.
  • Jul 13, 2009, 07:21 AM
    thinay

    Ok! Ok! I'll do that. :)

    Hey! :D What do you think in this problem? Is it positive or negative infinity?
    lim (x/x+2) + (1/x^2-4)
    x->-2^+

    when I tried to answer it. I end with + infinity.
  • Jul 13, 2009, 07:26 AM
    thinay

    My answer is wr0ng! :DD
    It should be - infinity..
  • Jul 13, 2009, 07:28 AM
    Unknown008

    Wait a sec. Does that mean you and akotoh are the same person? I've seen the exact question on another recent thread :confused:

    Anyway, the question just got answered.
  • Jul 13, 2009, 07:40 AM
    thinay

    Nope! I have read his/her post and I'm just confused what really the answer is. Because the other one says that the answer is - and the other says that its +.
    That's it! :)
  • Jul 13, 2009, 07:46 AM
    Unknown008

    Just had the graph plotted. It goes to -ve infinity, with asymptote at x = -2.
  • Jul 13, 2009, 07:58 AM
    thinay

    Hey! I've tried to substitute the value of x to g(x)= x/(x-5)..

    if..
    g(6) = 6
    g(7) = 3.5
    g(8) = 2.67
    g(9) = 2.25
    g(10) = 2

    am I correct? :D
  • Jul 13, 2009, 08:00 AM
    Unknown008

    Yup! :)
  • Jul 13, 2009, 08:05 AM
    thinay

    Okey! I'm totally done! :D
    Thank you very much for your help!

    I need to sleep now. Its already 11pm here in our country.

    Hope you can help me again next time! Thanks! More powers! ;)
  • Jul 13, 2009, 08:07 AM
    Unknown008

    Ok, Good night! :)

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