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Thread: Find a limit

  1. #1
    New Member Viona's Avatar
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    Find a limit

    Hello every one,

    I need help to understand how this limit is evaluated:

    lim.jpg
    I found it in a physics text book but I could not understand how they did it.

    Thanks

  2. #2
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    Quote Originally Posted by Viona View Post
    Hello every one,

    I need help to understand how this limit is evaluated:

    lim.jpg
    I found it in a physics text book but I could not understand how they did it.

    Thanks
    First, please read this summary of what we do, how we do it, and what you must do to get help.

    https://www.freemathhelp.com/forum/t...l=1#post433156

    In this case, we particularly need to know whether you have studied limits in a calculus or a pre-calculus course.

    Second, we need more context because what you have shown us is odd. Perhaps there is more that you have not shown.

    Third, please tell us what you have tried or thought while trying to understand.

    EDIT: What you have been shown is not math, but intuition.

    If x is very large, then

    [tex]\dfrac{2(x + a + 1) - b}{x} = \dfrac{2x}{x} + \dfrac{2a + 2 - b}{x} \approx 2 + 0 = 2.[/tex]

    Similarly, if x is very large,

    [tex]\dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.[/tex]

    So if x is very large

    [tex]\dfrac{2(x + a + 1) - b}{(x + 1)(x + 2a + 2} = \dfrac{\dfrac{2(x + a + 1) - b}{x}}{ (x + 1) * \dfrac{x + 2a + 2}{x}} \approx \dfrac{2}{(x + 1) * 1} = \dfrac{2}{x + 1}.[/tex]

    This is kind of foolish however because if x is very large, the resulting approximation is itself is approximately zero.
    Last edited by JeffM; 12-07-2018 at 10:47 AM.

  3. #3
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    Quote Originally Posted by Viona View Post
    Hello every one,

    I need help to understand how this limit is evaluated:

    lim.jpg
    I found it in a physics text book but I could not understand how they did it.

    Thanks
    Clearly this is not actually a limit, since x can't be in the result. It is something more like an asymptotic approximation.

    Please quote what was said leading up to, or explaining, this.

  4. #4
    New Member Viona's Avatar
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    Quote Originally Posted by JeffM View Post
    First, please read this summary of what we do, how we do it, and what you must do to get help.

    https://www.freemathhelp.com/forum/t...l=1#post433156

    In this case, we particularly need to know whether you have studied limits in a calculus or a pre-calculus course.

    Second, we need more context because what you have shown us is odd. Perhaps there is more that you have not shown.

    Third, please tell us what you have tried or thought while trying to understand.

    EDIT: What you have been shown is not math, but intuition.

    If x is very large, then

    [tex]\dfrac{2(x + a + 1) - b}{x} = \dfrac{2x}{x} + \dfrac{2a + 2 - b}{x} \approx 2 + 0 = 2.[/tex]

    Similarly, if x is very large,

    [tex]\dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.[/tex]

    So if x is very large

    [tex]\dfrac{2(x + a + 1) - b}{(x + 1)(x + 2a + 2} = \dfrac{\dfrac{2(x + a + 1) - b}{x}}{ (x + 1) * \dfrac{x + 2a + 2}{x}} \approx \dfrac{2}{(x + 1) * 1} = \dfrac{2}{x + 1}.[/tex]

    This is kind of foolish however because if x is very large, the resulting approximation is itself is approximately zero.

    Yes I studied limits long time ago, but this one seemed peculiar to me. Now I think it is some kind of asymptotic behavior. Thank you

  5. #5
    New Member Viona's Avatar
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    Quote Originally Posted by Dr.Peterson View Post
    Clearly this is not actually a limit, since x can't be in the result. It is something more like an asymptotic approximation.

    Please quote what was said leading up to, or explaining, this.
    I think you are write. I misunderstood what they mean. This actually what the book said:
    rec.jpgThanks.

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