Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Evaluating Limits

  1. Sep 5, 2011 #1
    Why is it that lim n->inf sin(n)^(1/n) = 1? I can't seem to figure it out why, thanks for any help that you can provide.
     
  2. jcsd
  3. Sep 5, 2011 #2
    What kind of values would Sin(n) take where whatever (n) is ?
     
  4. Sep 5, 2011 #3
    All values from -1 to 1, including these values, and then you would raise all those values to ^(1/n), were n is all values from 1 to infinity including 1, still unsure why though
     
    Last edited: Sep 5, 2011
  5. Sep 5, 2011 #4
  6. Sep 5, 2011 #5

    lanedance

    User Avatar
    Homework Helper

    its an interesting one

    now do you need to prove it or just convince yourself qualitatively?

    first what is the limit of
    [tex] \lim_{n \to \infty}(-1)^{1/n}[/tex]

    in fact for any number a, not zero, what is
    [tex] \Lim_{n \to \infty}(a)^{1/n}[/tex]

    another thing that is instructive to do might be to consider the magnitude in the limit

    The problems with sin will be around the zero, or n = k.pi, for integers k. For large n the exponent will "squeeze" these points of divergence to an infinitesimal region
     
  7. Sep 6, 2011 #6

    dynamicsolo

    User Avatar
    Homework Helper

    It is correct to say that you will need the "Squeeze Theorem" in some way, because L'Hopital's Rule will be useless since [itex]\lim_{n \rightarrow \infty} \sin(n) [/itex] does not exist.

    However, you will need to avoid using the problematic function [itex]a^{1/n}[/itex] , with a < 0 : it has a countably infinite number of undefined points, those being where n is even. (I notice, in fact, that Grapher won't even attempt y = (-1)1/x .) The proof will need to be drawn a bit more carefully...
     
  8. Sep 6, 2011 #7
    I originally thought that as well

    lim n->inf a^(1/n) = 1

    so

    lim n->inf sin(n)^(1/n) = 1

    but I wasn't so sure about that because lim n->inf sin(n) is undefined but would contain some value between -1<= sin(inf) <=1, at least I think

    and ya interestingly enough for some reason (-1)^(1/inf) is -1 and not positive one, not really sure why, (-1)^(1/inf) should be equal to (-1)^0 and should be equal to one 1 but it's not and actually negative one, not really sure why it's -1 and not 1, 0^(1/inf)=0^0 is undefined as well.
     
  9. Sep 6, 2011 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you mean lim n->inf sin(n)^(1/n) where n is a real number then to say it's 1 is silly. If n=k*pi, then sin(k*pi)=0 and 0^(1/n)=0. That's not near 1. If you mean n to be an integer then it's not so obviously wrong because k*pi is never an integer, but it can be REALLY CLOSE to an integer.
     
  10. Sep 6, 2011 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why would you think lim n->inf (-1)^(1/n)=(-1)?? First of all you have to DEFINE what (-1)^(1/n) means since there are n different nth roots. By making a sufficiently bizarre choice I think you can make the limit come out to any complex number z such that |z|=1. The usual principal root choice would make it +1.
     
  11. Sep 6, 2011 #10

    dynamicsolo

    User Avatar
    Homework Helper

    Actually, having thought about this general term a bit more and making a graph of it, I think the correct answer is that this limit does not exist. The function [itex]f(x) = (\sin x)^{1/x}[/itex] is undefined exactly half the time , which means that it has no meaningful value at half of the integers n .

    In the intervals where this function is defined, it ranges from 0 to 1 , with the associated curve coming to resemble increasingly a rectangle of unit height. The function is still only equal to 1 at odd multiples of pi/2 and becomes closer and closer to that value over a broader section of the defined intervals, but I don't think one can consider that to define a limit. (Try doing a classical "epsilon-N" proof on this critter...)

    I will again point to the deficiencies of computer algebra/calculus systems in the fact that Wolfram Alpha actually claims a limit at infinity for this function. You really shouldn't take the word of a "chunk of code" too seriously...
     
  12. Sep 6, 2011 #11

    lanedance

    User Avatar
    Homework Helper

    I think if you make some assumptions you can start to justify a limit,

    so following on from Dick and trying to deal with points on multivalued functions, if you put in the following caveats:
    - consider finding the limit of the series: sin(n)^(1/n) for n = 1,2,3,...
    - as exponentiating to a fraction is in general a multivalued operation, choose the principal root

    the fact any integer n is never equal to k.pi removes the issues with zeroes of sin(n)
    choosing the principal root means removing the issues with a multi-valued function

    in the previous post I was not laying out a blue-print for a comprehensive proof, but asking the OP what they wanted to achieve, and hopefully giving some insights that should help find them to a way to approach the problem

    so for the limit
    [tex] \lim_{n \to \infty }(-1)^\frac{1}{n}[/tex]

    why not consider [itex] -1 = e^{i \pi}[/itex], whilst acknowledging this is "choosing" a branch when the operation is multi-valued , which I'll return to later.
    [tex]\lim_{n\to \infty }(-1)^\frac{1}{n}
    =\lim_{n \to \infty }(e^{i \pi})^\frac{1}{n}
    =\lim_{n \to \infty }e^{i\frac{\pi}{n}}
    [/tex]

    in this form the limit becomes clear

    now in terms of the multivalued function, for the case [itex] -1^{\frac{1}{n}}[/itex] it is true we could choose [itex] -1 = e^{i (2k-1) \pi}[/itex] for k = 1,2,3... which gives [tex]\lim_{n\to \infty }(-1)^\frac{1}{n}
    =\lim_{n \to \infty }(e^{i (2k-1) \pi})^\frac{1}{n}
    =\lim_{n \to \infty }e^{i\frac{(2k-1) \pi}{n}} = 1, \ for \ \ finite \ \ k
    [/tex]

    It only really makes sense choosing k such that 0<(2k-1)<n, and if the series starts from 1, the natural choice is probably k=1. And for any choice of finite k the limit still tends to 1.

    In fact if you think of it as a function [itex] f(x) = (-1)^{\frac{1}{x}}[/itex] then changing this k value would in fact introduce discontinuities into what is a continuous function with f(1) = -1. That forces us to set a finite k value, in effect choosing a single-valued branch of the function. All the paths have f(1) = -1, and appear to tend to 1 for large n.

    If you consider plotting the function in the complex plane along the x axis, in a pseudo-3D sense, then I think they are are different "stretched spirals" in the complex plane with starting at f(1) = -1 converging on 1 for large x, with |f(x)|=1. The number of half revolutions of the x-axis is (2k-1).

    All that said its been a while since I've played with multi-valued functions, so feel free to look for any mis-assumptions/inconsistencies.

    Now if you consider that n is an integer, then we know sin(n) is non-zero and for any n we for some real a, with 0<a<1 we can can write:

    [tex] sin(n) = a [/tex]
    or
    [tex] sin(n) = a(-1) [/tex]
     
    Last edited: Sep 6, 2011
  13. Sep 6, 2011 #12
    As n approaches infinity 1/n approaches 0 and anything to the 0 is equal to 1
     
  14. Sep 6, 2011 #13

    dynamicsolo

    User Avatar
    Homework Helper

    Well, not just anything: if the base "grows as fast" as the exponent is shrinking, you can get a constant that isn't 1 . Example: [itex]lim_{x \rightarrow \infty} (1 + e^{x})^{1/x} = e [/itex] . If the base "grows faster", the function can run off to infinity -- for instance, [itex]lim_{x \rightarrow \infty} (1 + e^{x^{2}})^{1/x} = +\infty [/itex].
     
  15. Sep 7, 2011 #14
    Ah, thankyou :) Sorry about that GreenPrint
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook