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

Limits and infinite oscillations

  1. May 16, 2009 #1
    Hello everyone,

    I am having trouble understanding the concept of a limit not existing for functions like sin (1/x) when x tends to 0. The good book says that the function "does not settle on any value as we get closer to x" implying some infinite oscillation. I am having trouble visualizing it and why it should happen with this particular function.

    Any kind soul here willing to elaborate on this and help me understand this better? I would be extremely grateful.

    Many thanks,

  2. jcsd
  3. May 16, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    Think about sin(x) as x goes to infinity. Do you know what the graph of sin(x) looks like? Does it settle down to any particular value as x goes to infinity?
  4. May 16, 2009 #3
    Hi there,

    Thanks for the reply. The sine graph is oscillating between -1 and 1 every with a period of 2 pi. How would this change into infinite oscillations for sin(x) as x approaches infinity...


  5. May 16, 2009 #4
    Ok, I think I understand this better now as we approach infinity, the value will keep oscillating between -1 and 1 and there is no one value where the function will settle down to...fair enough :)


  6. May 16, 2009 #5


    User Avatar
    Homework Helper
    Gold Member

    You said it yourself. It oscillates between 0 and 1. Just draw the graph of sin. As you go further and further in the positive direction of the x axis, does it settle down to some value? No, it keeps oscillating between 1 and 0.

    EDIT: Sorry, I don't know what I was thinking. I meant to say oscillating between 1 and -1.
    Last edited: May 16, 2009
  7. May 16, 2009 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    Another way to look at it:

    [itex]sin(n\pi)= 0[/itex] for all n. Let [itex]x_n= 1/(n\pi)[/itex]. Then as n goes to infinity, [itex]x_n[/itex] goes to 0 and [itex]sin(1/x_n)= 0[/itex] for all n.

    But [itex]sin(2n\pi+ \pi/2)= 1[/itex] for all n. Let [itex]x_n= 1/(2n\pi+ \pi/2)[/itex]. Then as n goes to infinity, [itex]x_n[/itex] goes to [itex]sin(1/x_n)= 1[/itex] for all n.

    If [itex]\lim_{x\rightarrow 0} sin(1/x)[/itex] existed, the limit of [itex]sin(x_n)[/itex] would have to be that value for any sequence [itex]x_n[/itex] converging to 0. Since the two sequences above have different limits, the limit of the function cannot exist.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Limits and infinite oscillations
  1. Infinite limit (Replies: 5)