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Question about limits and horizontal asymptotes

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the limit and any horizontal asymptotes:

    lim 4/(e-x)
    x→∞

    2. Relevant equations

    Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞.

    Principle 2: If a limit goes to ∞ or -∞, there won't be a horizontal asymptote.

    3. The attempt at a solution

    lim 4/(e-x) = ∞
    x→∞

    since as x gets bigger, ex also gets bigger.

    Horizontal Asymptote = 0

    Given the above answer, if correct, then I have a problem with the two above principles I learned. Should I email my professor about these two principles? I am trying to learn these first before I do the actual homework.

    Thanks!
     
    Last edited: Sep 7, 2013
  2. jcsd
  3. Sep 7, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    So this is your function if I'm not mistaken : ##\frac{4}{e^{-x}} = 4e^x##.

    As ##x → ∞##, ##4e^x → ∞##. So there are no horizontal asymptotes happening in this direction.

    Why do you think ##y=0## is a horizontal asymptote for your function? Can you elaborate?
     
  4. Sep 7, 2013 #3
    Hi Zondrina!

    Yes, you interpreted the formula correctly.

    I didn't realize horizontal asymptotes were subject to a direction. I thought they applied to the whole function, so I saw that as x→-∞, the function approached 0. So I thought that was a horizontal asymptote (y = 0) for the function.

    EDIT: I will also go back and edit the function I posted in my first post to remove the ambiguity! I have to learn how to use the LaTeX function on this forum, but it was complicated when I first tried to even write a simple fraction!
     
  5. Sep 7, 2013 #4

    Zondrina

    User Avatar
    Homework Helper

    When you went to quote my second post, it will allow you to see how I wrote the latex. You can use that to get a feel for it. Here's a good tutorial to teach you some basics as well ( complements of micromass ) :

    https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

    Also you are correct when you say this :

    I was just checking if you explicitly knew to take ##x→-∞## instead. You have to check both directions with these kinds of problems.
     
  6. Sep 7, 2013 #5
    Thank you!

    So I hope this is right:

    as x→∞, the limit is also ∞, and there is no horizontal asymptote. But as x→-∞, both the limit and horizontal asymptote is 0.

    For the overall function, the horizontal asymptote is y = 0.

    And I also will go back and check the LaTeX. I would love to learn to write fractions and radicals, etc., correctly so they are easier to read for everyone involved. I have just been scared to try something new.
     
  7. Sep 7, 2013 #6
    Also, technically, Principle 1 above:

    "Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞."

    is violated here as x→∞, since the limit of the function here is ∞ but there is no horizontal asymptote.

    Maybe I should email my professor and ask her to clarify this. It worked for the examples she gave in class when she said the principle:

    lim(3x3 - 1000x2)
    x→∞

    lim(-x5 - x2 + x - 10)
    x→∞
     
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