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!

Homework Help: Continuety parameter question

  1. Jan 18, 2009 #1
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 18, 2009 #2
    Re: part A: I think this is a "setup" to make you think the answer is easy. Yes, it appears that: for m =1,2,3,4 ... etc, the function is continuous at x=0. Your logic looks good to me.

    However, the sin(1/x) function oscillates infinitely fast as x --> 0, so I think you'll find the answers to B and C to be surprising.
  4. Jan 18, 2009 #3
    how to solve it?
  5. Jan 18, 2009 #4
    Can you differentiate f(x) "formally" ? I get this:

    f'(x)= mxm-1sin(1/x) + xm-1cos(1/x)(-1)x-2

    For the case of m=1, I get:

    f'(x) = sin(1/x) - cos(1/x)/x2

    The first term doesn't approach anything as x--> 0. It oscillates infinitely fast between +1 and -1. No limit, therefore not continuous.

    Also, in second term, cos(1/x) -->1, but 1/x2 --> infinity, so again no limit.

    Your assignment, Mr. Phelps, should you choose to accept it ... :) ... is to find an m > 0 where f'(0) is continuous. See if you can find one.
  6. Jan 19, 2009 #5
    i got a solution to A
    and it says
    some thing about

    if m=0 then

    x_n = \frac{1}{(2n)\pi}
    i dont know how they came up with this Xn
    and what i need to do with it
  7. Jan 19, 2009 #6
    Oops, I missed the m > OR = 0 part.
    For f(x) = xmsin(1/x)
    If m=0, then f(x) = sin(1/x)
    So, for m=0, your logic for testing continuity of f(x) at zero breaks down. Sin(1/x) is not continuous at 0.

    Now, regarding that "solution" you found that says:
    [itex]x_n = \frac{1}{(2n)\pi} [/itex]
    Hmm. Well, let's look at when sin(1/x) = 0
    then 1/x = {[itex]\pi[/itex], 2[itex]\pi[/itex], 3[itex]\pi[/itex], ... } = n [itex]\pi[/itex]
    So I get: x = [itex]\frac{1}{(n)\pi} [/itex]
    Different than your solution, but the same idea: as n gets larger, x gets closer to 0, but ... between every point where sin(1/x) = 0, it oscillates up to +1 (or down to -1):

    when sin(1/x) = 1 --> 1/x = {[itex]\pi[/itex]/2, [itex]2\pi[/itex]+[itex]\pi[/itex]/2, ...} = [itex]2n\pi[/itex]+[itex]\pi[/itex]/2 --> x = [itex]\frac{1}{(2n\pi + \pi/2)}[/itex]
    and when sin(1/x) = -1 --> 1/x = {[itex]3\pi[/itex]/2, [itex]2\pi[/itex]+[itex]3\pi[/itex]/2, ...} = [itex]2n\pi[/itex]+[itex]3\pi[/itex]/2 --> x = [itex]\frac{1}{(2n\pi + 3\pi/2)}[/itex]

    So ... as x ->0, it passes through a full oscillation between each successive [itex]x_n = \frac{1}{(2n)\pi} [/itex], taking on values 0, 1, 0, -1, 0 successively between xn and xn+1

    That's a long-winded answer. Does it help?

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