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Indeterminate Forms and l'Hopital's Rule

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)


    2. Relevant equations



    3. The attempt at a solution

    When I initially plugged in ∞ for my x, I get (∞/∞)^∞, correct?

    If so, should I just let y=((2x+1)/(2x-1))^(sqrtx) and take the limit of both sides using ln?

    That's what I attempted to do and I got lim x→∞ of (sqrtx)(ln((2x+1)/(2x-1))^(sqrtx)) = (inf)(ln(∞/∞)) which I can't make any sense of.
     
    Last edited: Nov 6, 2012
  2. jcsd
  3. Nov 6, 2012 #2

    SammyS

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    Hello Reefy. Welcome to PF !

    I'm not sure what you mean by: "take the limit of both sides using ln?"

    You should consider taking the natural log of both sides of
    [itex]\displaystyle
    y=\left(\frac{2x+1}{2x-1}\right)^\sqrt{x}\ .[/itex]​
    Then see if you can get a suitable indeterminate form so that you can use L'Hôpital's rule .
     
  4. Nov 6, 2012 #3
    Thanks, I've been lurking for awhile and decided to make an account.

    What I meant is exactly what you said lol. I wrote:

    lim as x→∞ of lny = lim as x→∞ of ln((2x+1)/(2x-1))^(√x)

    When I brought the √x down I got, lim as x→∞ of (√x)(ln((2x+1)/(2x-1)) = (∞) times ln(∞/∞).

    Should I be getting ln (∞/∞) or am I doing something wrong? If it's correct, I don't know what to do from there on

    Edit: How do I write out the equations and functions like you did so it can be clearer?
     
  5. Nov 6, 2012 #4

    Mark44

    Staff: Mentor

    No. Your answer can't be ∞/∞, as that is indeterminate. Now, since (2x + 1)/(2x -1) → 2 as x → ∞, what you really have is [1], which is another indeterminate form.
    That should be lim(x→∞) [√x * ln((2x+1)/(2x-1))]. Keep in mind that the expression here is different from what you started with - it's the log of that expression. Any limit you get will be the log of what you should have for the original expression.
     
  6. Nov 6, 2012 #5

    SammyS

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    Last question first:

    Those mathematical expressions are written using something called LaTeX. Here are some links which may help you with LaTeX.

    https://www.physicsforums.com/showthread.php?t=617570&highlight=latex

    https://www.physicsforums.com/misc/howtolatex.pdf

    https://www.physicsforums.com/showthread.php?t=514469

    To get to the online LaTeX editor, click the Big Ʃ icon above the "Advanced" message box -- the box in which you compose your posts.

    Now for the main subject. (I see that Mark44 has answered your post while I was composing the above.) I'll just make one point.

    Write [itex]\displaystyle \ \ \ln\left(\frac{2x+1}{2x-1}\right)\ \ \ \text{ as }\ \ \ \ln(2x+1)-\ln(2x-1)\ .[/itex]
     
  7. Nov 6, 2012 #6
    Oops, I meant to take out that √x from the exponent. Thanks for pointing that out. I also tried the other way you showed with 1^∞ but didn't know what to do.

    Oh, I forgot that I can rewrite the natural log. Lemme see if I can get the write answer now.
     
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