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Limit of 0/0 indeterminant form, n and k for existence of limit

  1. Jun 21, 2012 #1

    AGNuke

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    If [tex]\lim_{x \to 0}\frac{e^{nx}+e^{-nx}-2cos\frac{nx}{2} - kx^2}{sinx - tanx}[/tex] exists and finite, then possible values of 'n' and 'k'



    2. Relevant equations
    By far, I have got one equation relating n and k.
    [tex]\frac{5n^2}{4}-k=0[/tex]

    I can simply put choices in there and get answer, but it would be better if I can get answer from the question itself.

    3. The attempt at a solution
    By removing the indeterminancy in the denominator and expanding the numerator, I got

    [tex]\frac{\left [2\left ( \frac{(nx)^2}{2!}+\frac{(nx)^4}{4!}+... \right )-2\left ( -\frac{(nx/2)^2}{2!}+\frac{(nx)^4}{4!}+... \right ) \right ]cosx}{-\left ( \frac{sinx}{x} \right )\left ( \frac{1-cos^2x}{x^2} \right )x^3}[/tex]

    I expanded the series till x3 or x4. I just want an equation which can confirm the value of n, which is probably 2 (as I am getting a fancy answer, k = 5 :smile:)
     
    Last edited by a moderator: Jun 21, 2012
  2. jcsd
  3. Jun 21, 2012 #2

    SammyS

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    Have t\you tried L'Hôpital's rule ?
     
  4. Jun 21, 2012 #3

    Mark44

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    AGNuke, please don't use SIZE tags. I replaced your itex tags with tex tags.
     
  5. Jun 21, 2012 #4

    AGNuke

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    After applying the L'Hôpital's rule twice, I got

    [tex]\frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}cos\frac{nx}{2}-2k}{-sinx - 2tanx.sex^2x}[/tex]

    After entering x = 0; since the denominator approaches zero, so must the numerator. After doing so, I got that same previous equation

    [tex]\frac{5n^2}{4}-k=0[/tex]

    OK and Thanks. I don't have much experience with latex and their size issue.
     
  6. Jun 21, 2012 #5
    Isn't that still an indeterminate form? Even if both the denominator and numerator approach 0 as x tends to zero, it doesnt guarantee that the limit exists, for one may reach 0 faster than the other.
     
  7. Jun 22, 2012 #6

    AGNuke

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    But the question states that the limit DO exist. That's why I made both of them to approach 0 as the denominator is going to do that for sure.

    And I tried using n = 2 and eventually k = 5, the limit is 0.
     
  8. Jun 22, 2012 #7
    Ah, I get your point. It looks like there are infinitely many solutions though.
     
  9. Jun 22, 2012 #8
    Assuming you haven't solved this yet, L' Hospitals's rule looks REALLY tempting here.
     
  10. Jun 22, 2012 #9
    Thats like saying the limit doesn't exist.
     
  11. Jun 22, 2012 #10

    SammyS

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    That all looks good.

    [itex]\displaystyle \frac{5n^2}{4}-k=0\quad\to\quad k=\frac{5n^2}{4}[/itex]

    So plug-in [itex]\displaystyle \frac{5n^2}{4}[/itex] for k :

    [itex]\displaystyle \frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}\cos\frac{nx}{2}-2k}{-\sin x -2\tan x\sec^2 x}
    \quad\to\quad \frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}\cos\frac{nx}{2}-\frac{5n^2}{2}}{-\sin x(1 +2\sec^3 x)}[/itex]

    and call on L'Hôpital one more time.
     
  12. Jun 22, 2012 #11

    AGNuke

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    If I do so, wouldn't the numerator becomes zero all on its own?
     
  13. Jun 22, 2012 #12

    SammyS

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    Yes.

    How about the denominator?
     
  14. Jun 22, 2012 #13

    AGNuke

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    Denominator was 0 right from the beginning.

    Even if I apply LH rule twice, I still won't get any respectable equation.
     
  15. Jun 22, 2012 #14

    SammyS

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    What's the derivative of [itex]-\sin x(1 +2\sec^3 x)\ ?[/itex]

    That derivative is not zero at x=0 .
     
  16. Jun 22, 2012 #15

    AGNuke

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    If I apply LH rule once after that fancy equation, maybe the denominator is -3, but the numerator becomes 0 all on its own, without leaving a gap for determining the value of n.
     
  17. Jun 22, 2012 #16

    SammyS

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    Then that means that the limit is 0, doesn't it?
     
  18. Jun 22, 2012 #17
    Are you allowed to use infinitesimal of equivalent?
    Note that 3(sin(x)-x) and sin(x)-tan(x) are equivalent infinitesimals as x approaches 0. Which makes much less miserable to apply L'Hôpital's rule multiple times.
     
  19. Jun 23, 2012 #18

    AGNuke

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    Yeah, It was obvious from the start, since that relation between n and k was established assuming limit is 0.
     
  20. Jun 23, 2012 #19

    AGNuke

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    Hmm... But my problem is not denominator, it's numerator.

    What I actually need is to find the value of n so I can find k as well using the relation derived earlier.

    For now, I answered the question through glamorous "hit-and-trial" considering it works best if this problem is to be solved under two minutes. Cancelling 4 with n2 with n = 2 and getting k = 5. (And the limit do exist, 0)
     
  21. Jun 23, 2012 #20

    SammyS

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    You got that relation between n and k, by making the limit of the numerator be zero, not by making the limit of the overall expression be zero.

    There's a BIG difference.
     
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