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It seems easy but I'm confused - limit

  1. Jan 16, 2005 #1
    Hi,

    I've been trying to compute

    [tex]
    \lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)
    [/tex]

    But don't have any idea. According to the results it should be 0, but I can't see the way I should get it...

    Thank you.
     
  2. jcsd
  3. Jan 16, 2005 #2
    for x>>1.. the square root approches to an integer... (even if it is going to infinite)
     
  4. Jan 16, 2005 #3
    Yes, the square root approaches to infinity I think, but I still don't get it.
     
  5. Jan 16, 2005 #4
    it is also an integer... sin(2pi n) =0
     
  6. Jan 16, 2005 #5
    Well, that's the thought I got instantly after the first look, but I'm trying to produce some serious proof of that...If it were in test, do you think that your answer would be accepted?
     
  7. Jan 16, 2005 #6

    Curious3141

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    The limit should be undefined. Sine is periodic and there is no way to define what's going to happen when you take such a limit.

    If x is defined to be a "very large" integer, I can see an argument for the answer zero, but by default x should be taken as real, and there is no defined limit.
     
  8. Jan 16, 2005 #7

    Curious3141

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    That's only if you assume x to be an integer, and there's no basis for such an assumption.

    What if [itex]x^2 = e^2(10^{16}) - 1[/itex], for example ?
     
  9. Jan 16, 2005 #8
    because he said the result is zero.... it is reasonable for me to make an assumtion that x is integer.... otherwise.... obviously, the function is undefine
     
  10. Jan 16, 2005 #9

    Curious3141

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    But what 'results' is he talking about ? The natural tendency is to keep plugging in large integer values (like 10^16) into the calculator in which case of course, you'd get zero. It would also be wrong.
     
  11. Jan 16, 2005 #10

    Gokul43201

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    I second Curious,

    The function is well-defined, but the particular limit does not exist.

    Twoflower,

    How did you get these "results" ?
     
  12. Jan 16, 2005 #11
    He gave us these limits as recommended homework along with the results...The only that comes to my mind is that I miswrote the original limit and replaced "n" with "x" :-(
     
  13. Jan 16, 2005 #12

    dextercioby

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    Okay,if it was "n" instead of "x",then the limit still does not exist.Moreover,it can be even shown it cannot be zero.For it to be zero,it is necessary that
    [tex] \lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2} [/tex]

    Now u have an equation to solve:are there natural "n"-s for which
    [tex] \sqrt{n^{2}+1}=\frac{k}{2} [/tex]
    ,for arbitrary "k",very large???
    My answer is "no",therefore the limit cannot be "0".Moreover,it doesn't exist.

    Daniel.
     
  14. Jan 16, 2005 #13
    Thank you Daniel but I can't see what couldn't the whole expression in sin be natural multiple of [itex]\pi[/itex] (and ie. [itex]\sqrt{n^2+1}[/itex] any integer)...
     
  15. Jan 16, 2005 #14

    dextercioby

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    Well,solve this equation
    [tex] \sqrt{n^{2}+1}=k [/tex]
    for natural "k" and "n".And pick out the (possible,if any) solutions,the ones for very large"n".

    Daniel.
     
  16. Jan 16, 2005 #15
    you confused the concept of limit again.... [tex] \sqrt{n^{2}+1}[/tex] approach to an integer while n is large... but [tex] \sqrt{n^{2}+1}[/tex] itself is not an integer.... proof:
    [tex] \lim_{n \rightarrow \infty} \sqrt{n^{2}+1} = n [/tex] where n is an integer... done
    [tex]
    \lim_{n \rightarrow \infty} \sin \left( 2\pi \sqrt{n^2+1} \right) = 0
    [/tex]
    sure, I have made an assumtion that n is integer...for an abitrary real number, the limit doesn't exist....
     
    Last edited: Jan 16, 2005
  17. Jan 16, 2005 #16

    dextercioby

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    So what's your point??Is that limit zero,or it doesn't exist?? :confused:

    Daniel.
     
  18. Jan 17, 2005 #17
    if n is integer, the limit exist, if not,then not....
    but you do agree with me that [itex] \lim_{x \rightarrow \infty} \sqrt{x^2+1} [/itex] is integer, right?
     
  19. Jan 17, 2005 #18
    Yes, this seems more logical to me than the Daniel's one...
     
  20. Jan 17, 2005 #19

    Hurkyl

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    No. The limit doesn't exist, so it can't be an integer. :tongue2:

    However, [itex] \lim_{x \rightarrow \infty} \mathrm{frac} \sqrt{x^2+1} = 0[/itex]

    If x is only allowed to range over integers. (frac is the fractional part, so frac 1.2 = .2)
     
  21. Jan 17, 2005 #20

    dextercioby

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    If [tex] +\infty [/tex] is an integer,i agree.

    Can u show that [tex] +\infty [/tex] is an integer????

    Daniel.
     
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