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Failure of L' Hopital's Rule

  1. Apr 20, 2005 #1
    Why does L' Hopital's fail for this limit? Both the numerator and the denominator are continuous, and both are differentiable.

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




    Is it "legal" to square the numerator and the denominator to get this:

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

    Now if i take this [tex]\lim_{x\rightarrow\infty}\frac{x^2}{x^2 + 1}[/tex] and use long divison, i get the following:


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

    and this limit is one. :devil:
     
  2. jcsd
  3. Apr 21, 2005 #2

    OlderDan

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    I'm not understanding your point. What is wrong with the limit being 1? Why do you need to square it fo find the limit? As for it being legel, what would squaring do if the limit were something other than 1? Where did you apply L'Hospital's rule?
     
  4. Apr 21, 2005 #3
    yah, you never even applied the rule, once you do you ultimately get sqrt(x^2+1)/x, and I believe you can just ignore the details and know that the x^2 on top is under the square root, so it's gonna ultimatly be to the power of 1, same with below. Thus 1
     
  5. Apr 21, 2005 #4

    OlderDan

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    OK.. after a quick google I see this is a known problem. I guess we are all supposed to know about the oscillations of the ratios of the derivatives. If you take enough of them you get back to where you started. Not being a pure mathemetician, I can speak what I suppose might be considered heresy. That's what you get for applying a rule for "indeterminate forms" to things that have an obvious limit. :rolleyes:
     
  6. Apr 21, 2005 #5
    oh right, oscillations of the derivatives. Yah, I was toootaly about to guess that. :uhh:
     
  7. Apr 21, 2005 #6

    i didn't use L' Hopital's Rule in my post. i used algebra to find the limit.
    this problem was in my problem set under L's rule. So when i did it using L's rule, L's rule failed.
     
  8. Apr 21, 2005 #7

    saltydog

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    One has to "prove" that the limit is 1. I think RadiationX is correct if he does the following:

    Let:

    [tex]y=\frac{x}{\sqrt{x^2 + 1 }}[/tex]

    So:

    [tex]y^2=\frac{x^2}{x^2 + 1}[/tex]

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

    Thus:

    [tex]\lim_{x\rightarrow\infty}y=\sqrt{1}=\pm 1[/tex]

    And since y is always positive for x>0, we choose +1.

    However, I'm open to a better way of "proving" it.
     
  9. Apr 21, 2005 #8

    StatusX

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    Just multiply the top and bottom by 1/x.
     
  10. Apr 21, 2005 #9

    Curious3141

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    I completely fail to see the issue the OP is bringing up.

    Using LH rule :

    [tex]\lim_{x\rightarrow\infty}\frac{x}{\sqrt{x^2 + 1 }} = \lim_{x\rightarrow\infty}\frac{1}{x{(x^2 + 1)}^{-\frac{1}{2}}} = \lim_{x\rightarrow\infty}\frac{{(x^2 + 1)}^\frac{1}{2}}{x} = \lim_{x\rightarrow\infty}{(\frac{x^2 + 1}{x^2})}^\frac{1}{2} = \lim_{x\rightarrow\infty}{(1 + \frac{1}{x^2})}^\frac{1}{2} = 1[/tex]
     
  11. Apr 21, 2005 #10
    I think the L'Hospital Rule is not failing, simply for certain problems is not necessary since problems of this kind are resolvable by algebraic methods since they are pseudoindeterminate forms.

    Something like this is, too, when a limit that implies a functions like sines, cosines or exponentials, the Rule gets infinit steps.

    A professor say me that this rule is perilous, it's better to use an expansion of Taylor series of the functions and use the properties of series to get the limit, since it would get you to false conclusions if you don't know how to use it correctly.

    An you curious are using the LH plus some algebra. The LH is intended to eliminate the indetermination by succesive LH. With you first and last LH you have get another indeterminate form. If you are using LH you must apply it another time, in fact it's infinity process in this case and you will finally don't get the limit. But you can stop and obtain the limit by algebra.
     
    Last edited: Apr 21, 2005
  12. Apr 21, 2005 #11

    saltydog

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    Alright, I didn't see that. That completely resolves the issue for me anyway.
    Thanks.
     
  13. Apr 21, 2005 #12

    dextercioby

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    Since l'Hôpital's rule "fails" (doesn't give any result),i'd suggest to consider this elegant solution

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

    Make the substitution

    [tex] x=\cot t [/tex]

    The limit becomes

    [tex] L=\lim_{t\rightarrow 0}\frac{\frac{\cos t}{\sin t}}{\sqrt{\left(\frac{\cos t}{\sin t}\right)^{2}+1}}=...=\cos 0=1 [/tex]


    Daniel.
     
  14. Apr 21, 2005 #13

    Curious3141

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    Alright, that's fair enough. But LH rule hasn't "failed" in any way (as in : given a wrong result), it just hasn't helped toward a solution. There are limits where LH doesn't help, and this is one. We're in agreement here.
     
  15. Apr 21, 2005 #14

    dextercioby

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    And if you guys still wonder why those "x" and "sqrt" go in circles,here's the proof

    "Make the substitution [tex] x=\sinh t [/tex]"...

    Daniel.
     
  16. Apr 21, 2005 #15
    I never said LH has failed. In fact if you see what i have wrote I say that the only thing is that LH is not inteded for this limits. And dextercioby, I think that is not elegant. In mathematics and physics I have learnt that a elegant thing is some thing that is symmetric and has a simple structure compared of the large number of applications that has it. I can say what you have done is a complicated form to do that, but it's not bad, in fact many times we must do some things like that.
     
  17. Apr 21, 2005 #16

    Curious3141

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    You didn't. The original poster did. I am not disagreeing with you.
     
  18. Apr 21, 2005 #17
    Here is the original question as it appeared in my text: Calculus Seventh editon, by Larsen, Hostetler, and Edwards.

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

    (a) Find the limit analytically without trying to use LH Rule

    (b) Show that LH Rule fails

    (c) use a graphing utitlity to graph the function and approximate the limit from
    the graph.Compare the results with that in part (a)
     
  19. Apr 21, 2005 #18

    dextercioby

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    I provided an easy way to do it,other than extracting "x" from the denominaor...

    There's no failure,just a no-go result,to use a fancy term.

    No need to graph.

    Daniel.
     
  20. Apr 21, 2005 #19

    OlderDan

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    The "failure" is that by taking sucessively higher order derivatives you get no closer to the limit than you were with the original functions. In fact after two derivatives you are right back where you started.

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

    [tex]\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}= \lim_{x\rightarrow\infty}\frac{f'(x)}{g'(x)}= \lim_{x\rightarrow\infty}\frac{g(x)}{f(x)}[/tex]

    [tex]\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}= \lim_{x\rightarrow\infty}\frac{f''(x)}{g''(x)} = \lim_{x\rightarrow\infty}\frac{g'(x)}{f'(x)} = \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}[/tex]

    You can evaluate the limit anywhere along the way, because you can evaluate it directly from the original ratio.
     
    Last edited: Apr 21, 2005
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