1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is sin x when x tends to infinity?

  1. Jun 4, 2005 #1
    This question popped up in my head. What is sin x when x tends to infinity? Since sine is a preiodic functions which repeats itself, is the answer 1, -1 or 0 or something else altogether?
  2. jcsd
  3. Jun 4, 2005 #2
    It is undefined; there is no limit! Sin[x] does not converge to any value as x increases - it maintains its periodic character. Limits only apply when something converges to something somewhere.

    A proof is readily constructed from the def. of limits at [tex]\mbox{$\infty$}[/tex]. Choose a small [tex]\mbox{$\epsilon$}[/tex] ("1"is small enough) and show that for any x0, no matter how large, there exists an x>x0 such that


    Plug in [tex]\mbox{$\epsilon=1$}[/tex] and [tex]\mbox{$x=x_0+\frac{\pi}{2}$}[/tex].

    What you show is that is not convergent, thus there is no limit.

    Hey, does anyone know a better way to do inline LaTeX here?
    Last edited by a moderator: Jun 5, 2005
  4. Jun 5, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    It looks good to me until [itex]\geq[/itex].

  5. Jun 5, 2005 #4
    What's wrong with [itex]\geq[/itex]? You want to show that [itex]\|f(x)-f(x_0)\|<\epsilon[/itex] does not hold for all [itex]x>x_0[/itex], so you find an [itex]x[/itex] where the relation is [itex]\geq[/itex] instead of [itex]<[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook