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Limits : the answer is not obvious!

  1. May 18, 2009 #1
    i have been racking my brains on this question for a long time and am still unable to reach to the answer.

    limit x → ∞ [(sin x + 2)/(sin x + 2) ] = ?

    limit x → ∞ [ (sin x + 0.5)/(sin x + 0.5) ] = ?

    the answers for both the questions are not the same.

    there's also a hint: think in terms of imaginary no.s and argand plane
     
    Last edited: May 18, 2009
  2. jcsd
  3. May 18, 2009 #2

    tiny-tim

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    Hi hermy! :smile:

    (have an infinity: ∞ :wink:)

    I don't think it has anything to do with complex numbers …

    it's more a question of whether the formula is defined for all values of x :wink:
     
  4. May 18, 2009 #3

    since sin x can never be 2, the limit for sin x + 2........ must be 1.

    but what puzzles me is why the limit shouldnt be 1 in the second case too?
     
  5. May 18, 2009 #4

    tiny-tim

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    Because sinx can be -0.5 :wink:
     
  6. May 18, 2009 #5
    Does it say that the limit for the second function is undefined?
     
  7. May 18, 2009 #6
    I don't get this. For any value of x other than arcsin(-.5), isn't the function identically 1? If so, doesn't that mean that as you approach those singular values from either side, it remains 1, and does not diverge anywhere? A naive application of L'Hopital's rule would seem to suggest the same thing.

    I would have said that both limits are identically 1, but I'm no mathematician.
     
  8. May 18, 2009 #7

    HallsofIvy

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    What do you mean by "either side"? I've heard of "To infinity and beyond" but I was under the impression that there was not an "other side" to infinity!

    Were you thinking of the limit as x goes to 0?
     
  9. May 18, 2009 #8
    I meant as you approach any of the singular points, i.e. the points where sin(x) = -0.5. Everywhere else it's well-behaved, right? So if it's equal to 1 at these points and everywhere else, why does it change as you go to infinity?

    In other words, I would have thought that both cases given would be equal in every way to
    [tex]lim_{x\rightarrow\infty}[/tex]1

    There must be some subtlety here that I'm missing.
     
    Last edited: May 18, 2009
  10. May 19, 2009 #9
    I was always of the impression that before you find limits, cancel anything that cancels, then rearange in order to get a form in which finding the limit is obvious... and in both of the limits you give it all cancels out. So either there is a typo, or you are asking for the limit as x tends to infinity of 1.

    however in responce to the imaginary hint, sin (x) can be written as;

    (eix-e-ix)/2i

    where i = [tex]\sqrt{-1}[/tex]
     
  11. May 19, 2009 #10

    Mark44

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    When you cancel, you are pulling out factors that are equal to 1. In the case of (sin x + .5)/(sin x + .5), that expression is equal to 1 as long as sin x is not equal to -.5.
     
  12. May 21, 2009 #11
    The answer to the first question is 1 and the second limit is not defined. I have fair confidence in the authencity of the answer. (you could visit the website http://thepuzzlespot.blogspot.com/2008/08/perfect-in-limitshmmmm.html" [Broken] , where i found this question)

    In my opinion, the explanation that sinx can be -0.5 does not suffice since x tends to infinity. sin x can be anything between -1 and 1. Isn't there any other way to go about the question?
     
    Last edited by a moderator: May 4, 2017
  13. May 21, 2009 #12

    tiny-tim

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    Hi hermy! :smile:
    That's correct:

    the second expression is undefined again and again as you "approach infinity", so you can't find an N such that f(x) even exists for all x > N, let alone that it converges to something :wink:
     
  14. May 21, 2009 #13

    HallsofIvy

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    In order to be able to say that "[itex]\lim_{x\rightarrow a} f(x)= L[/itex]", there must be some (punctured) neighborhood of a such that f(x) is defined for all x in that neighborhood. A "neighborhood of infinity" is a set {x> X} for some number X. No matter how large X is, there will be points in {x> X} for which (sin(x)- 0.5)/(sin(x)- 0.5) is not defined.
     
  15. May 21, 2009 #14
    I don't understand the statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined".

    This just looks to me like one of those first-year calculus problems where you are asked to determine the value of a ratio as both numerator and denominator go to zero. Is it zero? One? Infinite? We apply L'Hopital's rule to find the answer.

    In this case, it seems unambiguous to me. First, however, may I restate the problem to refer instead to the ratio (sin(x)-1.0)/(sin(x)-1.0)? It's just easier to speak of the values x = n*pi/2 rather than arcsin(0.5).

    EDIT: My mistake - please substitute "(4n+1)*pi/2" for "n*pi/2" throughout ... I was a little too hasty in my typing. (n is an integer, of course).

    If I were asked what is the value of (sin(x)-1.0)/(sin(x)-1.0) for x -> n*pi/2, I would differentiate top and bottom, per L'Hopital, to get cox(x)/cos(x), and happily conclude that the answer is 1.

    Is that incorrect? If not, and (sin(x)-1.0)/(sin(x)-1.0) is in fact equal to 1 for x = n*pi/2, then it is in fact equal to 1 for all x, in which case the limit as x goes to infinity is clear.

    What's the subtlety that I'm missing? Does it have to do with the difference between asking the value of (sin(x)-1.0)/(sin(x)-1.0) as x approaches n*pi/2 versus the value for x = n*pi/2 identically? I admit that I don't generally distinguish between the two in practice.
     
    Last edited: May 21, 2009
  16. May 21, 2009 #15

    jbunniii

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    The limit is being taken as x goes to INFINITY, not to arcsin(0.5). Neither the numerator nor the denominator has a limit as x goes to infinity, so L'Hopital's rule does not apply.

    The statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined" is true for any x that makes the denominator zero, and that happens with period 2*pi all the way out to infinity. That is why the limit does not exist.

    Of course, if the domain were restricted to values of x that don't make the denominator zero, or if the function were explicitly defined to be 1 at those points, then the limit would exist, but neither appears to be the case.

    That said, I have no idea what the original poster's hint ("think in terms of imaginary no.s and argand plane") has to do with this problem.
     
  17. May 21, 2009 #16
    I wasn't referring to that limit - I was referring to x = arcsin(0.5), which is where the function was asserted to be undefined. I understand that the limit as x goes to infinity is ill-defined if the function is undefined at these points, but I'm asking why it is undefined at these points.
    Of course a denominator of zero is problematic, but if the numerator is also zero, then you have to ask about the rate at which each one goes to zero as x approaches arcsin(0.5) - that's the basis for L'Hopital's rule, as I understand it. If the numerator goes to zero faster, then the ratio gets smaller as x approaches the limit, so we say the ratio goes to zero; if the denominator goes to zero faster, the ratio goes to infinity, and if they both go to zero at the same rate (as is the case here, since they're in fact equal), then the ratio goes to 1 - or in this case, it maintains a constant value of 1.

    So my confusion persists: why do we have say that the ratio (sin(x)- 0.5)/(sin(x)- 0.5) is not defined for x = arcsin((0.5)? I would assert that the ratio is exactly equal to 1.0, for the reasons I just gave.
     
  18. May 21, 2009 #17

    tiny-tim

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    Because when x = arcsin((0.5),

    (sin(x)- 0.5)/(sin(x)- 0.5) = (0.5 - 0.5)/(0.5 - 0.5) = 0/0,

    which isn't defined.
     
  19. May 21, 2009 #18

    jbunniii

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    You have shown that the LIMIT of the ratio as x APPROACHES arcsin(0.5) is 1.0. That is not the same as saying that the ratio EQUALS 1.0 when x EQUALS arcsin(0.5). It doesn't.

    If the definition was modified as follows:

    [tex]f(x) = \left\{\begin{array}{ll}
    1 & \textrm{if sin(x) = 0.5} \\
    \frac{\sin(x) - 0.5}{\sin(x) - 0.5} & \textrm{otherwise} \end{array}[/tex]

    Then it would be defined everywhere (in fact would equal 1 everywhere), so the limit would be 1.
     
    Last edited: May 21, 2009
  20. May 21, 2009 #19
    Why would you not say the same thing about any other function to which you applied L'Hopital's Rule?

    I thought that was the whole point - to disambiguate between the general rules:

    0/anything = 0
    anything/itself = 1
    anything/0 = undefined
     
  21. May 21, 2009 #20
    I must be suffering from a very basic misconception, here.

    Suppose I asked you what is the value of cos^2(x)/(sin(x)-1) at the point x = pi/2? Would you say this is undefined because the denominator is zero? I would not.

    Two applications of L'Hopital's Rule show that this function does not diverge - in fact it goes to the value -2. You can easily plot it to see that. I've attached a plot of cos^2(x) (blue), sin(x)-1 (red), and cos^2(x)/(sin(x)-1) (magenta).

    I need you guys to explain to me why L'Hopital's rule exists if you insist that a function must be undefined wherever its denominator goes to zero (regardless of the behavior of the numerator).
     

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