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: Limit of an absolute sequence

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the sequence a_n = abs(sin(x))^(1/x)
    Find the lim a_n if it exists

    2. Relevant equations

    None. This is for my calc 2 class.

    3. The attempt at a solution

    We are studying the sandwich theorem, so I thought 0 < M^(1/x) < abs(sin(x))^(1/x) < 1^(1/x).
    (Because I assumed that sequences imply x = 1, 2, 3, 4 ..., so sin(x) never equals 0).

    Since M^(1/x) and 1 both tend to 1, I reasoned a_n must go to 1.
    Last edited: Apr 8, 2008
  2. jcsd
  3. Apr 8, 2008 #2
    Do you mean [tex]a_{n}=|sin(n)|^{\frac{1}{n}}[/tex]? Second, [tex]\sqrt[x]{1}[/tex] does not tend to 0 as x becomes large, nor does even [tex]\sqrt[x]{\frac{1}{2}}[/tex]. That doesn't make using 1 any less valid, you just made an incorrect assumption.
    Last edited: Apr 8, 2008
  4. Apr 8, 2008 #3
    sorry i meant that lim goes to 1. (fixed the typo in original post)
  5. Apr 8, 2008 #4
    You're almost there, but I'm not convinced. Can you prove the existence of an M and that [tex]\sqrt[x]{M}[/tex] goes to 1 as x becomes large? I'm not sure how much rigor is required in your class.
    Last edited: Apr 8, 2008
  6. Apr 8, 2008 #5
    I reasoned that M exists because the real numbers are dense.
    and you can prove M^(1/x) goes to 1 using the definition of the limit.
    Are there any holes in my argument?
  7. Apr 8, 2008 #6
    Well, it won't equal *exactly* 0, but doesn't the density of the reals imply that it gets arbitrarily close? What is [itex]\liminf_{n\to\infty}\left|\sin(n)\right|[/itex]?
  8. Apr 8, 2008 #7
    hmm yeah i thought that part of my argument was a bit shady.
    can anyone offer some insights?
  9. Apr 9, 2008 #8


    User Avatar
    Science Advisor

    The real numbers are dense in what?

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook