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Challenge problem

  1. Feb 16, 2008 #1
    a challenge problem in my book states why dosent lhopitals rule work on the limit as x goes to infinity of (x^2+sinx)/x^2. im stumped
  2. jcsd
  3. Feb 16, 2008 #2
    I don't know if it doesn't work, it just isn't useful since differentiating once does not get you any further ahead and differentiating twice is not valid because the second derivative of the denominator is not defined for any open interval containing infinity.


    To correctly way to solve the limit is expand the above as follows


    bound the value of sinx and then use the squeeze theorem to show the limit is equal to one.
  4. Feb 16, 2008 #3
    It is only valid to use L'Hopital's rule once on this function, as you have both numerator and denominator increasing without bound. After you get the expression 1 + [itex]\lim_{x\rightarrow\infty} \frac{\cos x}{2x}[/itex], the expression no longer conforms to the conditions that validate the use of L'Hopital's rule.
  5. Feb 17, 2008 #4


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    No, slider.

    One of the conditions for the applicability of Hospital's rule is that the limit of the fraction of the derivatives MUST exist.

    In this case, that limit does not exist, but the limit of our original expression exist NONETHELESS (equaling 1).

    Read John's link carefully.
  6. Feb 17, 2008 #5
    If you apply it once the limit still exists.
  7. Feb 17, 2008 #6


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    Which limit? The limit of the original problem certainly exists but arildno's point is that trying to apply L'Hopital's rule the derivative of the numerator is 2x+ cos(x) and that has no limit as x goes to infinity.
  8. Feb 17, 2008 #7
    Yes but the denominator is x^2 so you can still apply the rule once and the limit will still exist.
  9. Feb 17, 2008 #8


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    ?? Now I am starting to wonder if you know what "L'Hopital's" rule is? As I said before, the limit of the derivative of the numerator does not exist. It doesn't matter what the denominator is!
  10. Feb 17, 2008 #9
    Ah yes. I was asleep when I wrote that nonsense.
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