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L'Hopital ad infinitum

  1. Jun 30, 2013 #1
    So I've come across this formula that I derived. y(t) =v2t/√(v2t2+b2)

    I would like to solve the limit of t to infinity analytically. When I apply L'Hopital I get
    y = lim v2 / lim v2t/√(v2t2+b2)

    but as you can see I would have to apply L'Hopital rule an infinite amount of times, now I don't know if you say it becomes x/(x/(x/..))). with x= v2 whatever value that is.

    By inspection of a grapher I would say it's v , it also looks like v*sin(arctan(v/b*t)), which at t-> arctan -> pi/2 then sin() -> 1 so the answer is v. But, how do I know arctan(t) t->inf it pi/2 besides geometrically it makes sense.

    Any ideas about the infinite L'Hopital, or infinite divisions how something like that could be solved.
    Last edited: Jun 30, 2013
  2. jcsd
  3. Jun 30, 2013 #2

    Simon Bridge

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    You mean find ##L:## $$L=\lim_{t \rightarrow \infty}\frac{v^2t}{\sqrt{v^2 t^2 +b}}$$

    I'd be inclined to take the vt out from under the square root. ##b/v^2t^2\rightarrow 0##.

    Under L'Hopital, if you just keep repeating it, I suspect you'll and up finding L=f(L) ...and solve for L.
  4. Jun 30, 2013 #3


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    You mean that you get
    [tex]\lim_{t \to \infty} y = \frac{\lim_{t \to \infty} v^2}{ \lim_{t \to \infty} \left( \frac{v^2 t}{\sqrt{v^2 t^2 + b^2}} \right) }[/tex]
    (note I put the missing limit on the left hand side), i.e.
    [tex]\lim_{t \to \infty} y = \frac{\lim_{t \to \infty} v^2}{\lim_{t \to \infty} y} ?[/tex]

    In Simon's notation, that would be
    [tex]L = \frac{\ldots}{L}[/tex]
    except that you don't have to apply L'Hopital more than once.
  5. Jun 30, 2013 #4

    Simon Bridge

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    The same trick is often used in integration by parts.
    Y'know, I'd never heard of L'Hopital before I started at this forum. It would have been handy. But in this case, do you really need it? Oh well, never mind.

    "Remember children: there are no stupid questions, only stupid people."
    -- Mr Garrett (Southpark)​
  6. Jun 30, 2013 #5
    Ah yeah, didn't see that way.

    Oh sure, kind of seems silly now [tex]\lim_{t \to \infty} y =√(\lim_{t \to \infty} v^2)= v[/tex]

    Always good to know a fews ways to an answer.
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