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Lupitals law

  1. Aug 28, 2008 #1
    how do i solve this limit using luptials law

    x->(infinity) (x)*[(x^2+1)^-0.5]

    i seem to be going in circles, it is meant to be a simple few steps to the solution but i cant see it.
  2. jcsd
  3. Aug 28, 2008 #2


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    FYI: You're thinking of "L'Hopital's Rule" not "Lupitals law".

    Do you insist on using L'Hopital's rule? The limit is easily solved by re-arranging the denominator to cancel the x in the numerator.
  4. Aug 28, 2008 #3
    Divide the top and the bottom by x.
  5. Aug 28, 2008 #4


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    That's "l'Hôpital's rule", which (following the standard rules for written French) can also be written "l'Hospital's rule". It says that in a limit you can replace an indeterminate fraction with the ratio of the derivatives of the numerator and denominator.

    So the numerator is x and the denominator is sqrt(x^2+1). Can you find the derivatives of these?
  6. Aug 28, 2008 #5


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    As other's have said, that's "L'Hopital" or "L'Hospital". I don't see how anyone can point out to you where you went wrong if you dont' show what you ahve done. It looks pretty straightforward to me. What did you attempt.
  7. Aug 28, 2008 #6


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    … infinite loop …

    Hi HallsofIvy! :smile:

    But devanlevin has told us where he got stuck …
    … if you apply L'Hôpital's rule, you do go round in circles … you get stuck in an infinite loop where you keep having to solve the original problem all over again!

    Does anyone know a standard way out of this, still using L'Hôpital's rule (so not dividing by x :wink:)? :smile:
  8. Aug 28, 2008 #7

    Ben Niehoff

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    Yes, there is a simple way. Using l'Hopital's Rule, we can see that

    [tex]\lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x \rightarrow \infty} \frac{\sqrt{x^2+1}}{x}[/tex]

    Then one applies the properties of limits; i.e., suppose the limit

    [tex]M = \lim_{x \rightarrow a} f(x)[/tex]

    exists and is not zero. Then one can easily find

    [tex]\lim_{x \rightarrow a} \frac{1}{f(x)}[/tex]

    using the properties of limits. Then merely set these limits equal to each other (according the first equation, derived via l'Hopital's Rule), and solve. (Sorry, I don't want to give away too much to the person who asked the question).
  9. Aug 28, 2008 #8
    Can't you only do this if you know the limit exists and is non-zero? Which you must prove seperately?
  10. Aug 28, 2008 #9


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    Hi Ben Niehoff! :smile:

    Yes … that's nice :smile: … although it's an endless loop, you can break out of it by using that trick after the first step … thanks!

    (hmm … I hadn't actually noticed it was upside-down :redface: … shouldn't have done it in my head! :rolleyes:)
  11. Aug 28, 2008 #10

    Ben Niehoff

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    I used Maple to assist, because I hate taking derivatives under square roots.

    I agree with the previous poster that one must consider the existence of the limit in the first place. Most of the time, you can simply apply the properties of limits anyway, and if the limit does NOT exist, you'll run into some inconsistency. So if you can find an internally-consistent answer, you can conclude that the limit must exist. I don't know if this is rigorous, though.

    A simpler way involves using properties of limits at the start:

    [tex]\lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2+1}} = \sqrt{\lim_{x \rightarrow \infty} \frac{x^2}{x^2 + 1}}[/tex]

    Applying l'Hopital's Rule here leads directly to the answer, with no infinite regress.
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