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Easy Limit Problem

  1. Jun 5, 2005 #1
    [tex]\lim_{x\rightarrow -\infty\\} \frac{\sqrt{5x^2-2}}{x+3}[/tex]

    When I first looked at it i thought the top will increase to inf. at a rate of [tex]\sqrt5[/tex] and the bottom will increase to inf. at a rate of -1, thus the answer would be [tex]-\sqrt5[/tex]

    However when i do the algebra i get this...
    [tex]\lim_{x\rightarrow -\infty\\} \frac{\sqrt{5x^2-2}}{x+3}[/tex]

    [tex]\lim_{x\rightarrow -\infty\\} \frac{x\sqrt{5-2/x^2}}{x(1+3/x)}[/tex]

    [tex] \frac{\sqrt{5-0}}{1+0}=\sqrt5[/tex] but the answer is negative :confused:
  2. jcsd
  3. Jun 5, 2005 #2


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    A subtle error to be sure. Because, by convention, the result of the square root operation on real positive numbers is always positive, note that [tex]\sqrt{x^2} = x[/tex] only holds for positive real x. For negative real x, [tex]\sqrt{x^2} = -x[/tex]

    In general, for real x, [tex]\sqrt{x^2} = |x|[/tex]. This is simply because of the convention by which the square root is defined.

    Can you now see your error ?
  4. Jun 6, 2005 #3
    The x I factored out on the top, was a negative x?
  5. Jun 6, 2005 #4
    As pointed out at http://mathworld.wolfram.com/LHospitalsRule.html, this is one of those rare cases where L'Hospital's rule fails. The example given there as u goes to infinity is:

    where the limit is 1. Well, since u can increase beyond bound, it is hard to see how the limit would be different if the case was: [tex]\frac{u+3}{\sqrt{u^2+15}}[/tex], which suggests a start on the problem. However I see that you have the right idea! but, since a square is positive, we have a negative value in the numerator of the original problem as pointed out by Curious 341.
    Last edited: Jun 6, 2005
  6. Jun 6, 2005 #5


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    Yes, it should be replaced by [itex]-x[/itex].
  7. Jun 6, 2005 #6
    Thx for the help

    I dont know what L'Hospital's rule is, I may learn that though
  8. Jun 6, 2005 #7


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    It's not relevant to this question, but basically it enables you to sort out limits of quotients of functions that look indeterminate from mere inspection. The limit you posted is of a quotient, but it's pretty trivial to solve with mere inspection after a little algebra.

    Certain conditions have to be met by the functions before LH rule can be invoked, and LH rule doesn't always help. It never really *fails* (as in gives a wrong answer) though, as long as the conditions are met.

    Read more about it here.
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