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Homework Help: Evaluating a limit with l'hopital's rule

  1. Apr 13, 2008 #1
    It's been a while since I've evaluated limits, and I'm beginning to forget some of the techniques. A problem came up in physics which involved evaluating a limit of this particular form.

    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \to \infty} \left( \frac{x}{\sqrt{x^2+y^2}} \right)[/tex]

    2. Relevant equations

    L'hopital's rule would be my first guess at the proper approach.

    3. The attempt at a solution

    I know the limit will evaluate to be a 1. This could also be inferred from the fact that x is in the same degree in both the numerator and the denominator. But I tried to do it a bit more rigorously with l'Hopital's rule. The problem is that the expression inside the limit becomes circular with successive derivatives. That is, if we let [itex]f(x) = x[/itex] and [itex]g(x)=\sqrt{x^2+y^2}[/itex], we will find that:


    This will bring us no closer to finding the limit. Take the derivative of the numerator and the denominator again to find:


    Hey, we're back! L'Hopital took us for a spin and brought us back to the starting point.

    So my questions:

    1. What is the proper approach to evaluate the limit?

    2. For the sake of curiosity, are there any interesting observations to be made about the situation? Perhaps a name for the circular nature of the problem?

    Thanks in advance for any help.
  2. jcsd
  3. Apr 13, 2008 #2
    I mean technically you're in a correct form of infinity over infinity, but thinking broadly: I learned l'hopistal's rule in single variable calculus. You're writing g(x,y) as g(x) by mistake (at least I think so). I don't know if you need to use partials in three space as such, but I always remembered that limits were expressed as, say, x,y -> 0. The answer to your limit is probably different if y->0 vs. y-> infinity. Sorry for the ad-hoc suggestions, I'm not really sure!
  4. Apr 13, 2008 #3
    Oh, sorry, in the context of the problem, [itex]y[/itex] is a constant, so putting [itex]y^2[/itex] is a little misleading. A more accurate statement of the problem would then be:

    \lim_{x \to \infty} \left( \frac{x}{\sqrt{x^2+c}} \right)

    This doesn't change a lot. We still get those circular derivatives.
  5. Apr 13, 2008 #4


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    HINT: As x becomes large, one can negate the constant :wink:
  6. Apr 13, 2008 #5


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    Sure it does -- if L is the value of the original limit, then what are the limits of the left and right hand sides of that equation?

    The method you wanted to use in the beginning is one "proper" approach. Didn't you learn how to do it rigorously with polynomials? Won't the exact same method work here?
  7. Apr 13, 2008 #6


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    l'Hopital is not guaranteed to resolve a limit. In some cases, depending on how you write the indeterminant form, it can actually make things worse. You found one where it doesn't help. There is another way, divide numerator and denominator by x and move the 1/x inside the radical.
  8. Apr 13, 2008 #7


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    Good point. You can salvage something from l'Hopital. IF you assume it has a limit.
    Last edited: Apr 13, 2008
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