1. Not finding help here? Sign up for a free 30min 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!

Continuity of sin(1/x) on (0,1)

  1. Dec 20, 2007 #1
    1. The problem statement, all variables and given/known data
    how do you show that sin(1/x) is continuous on (0,1)? (i know it's also continous on (0, infinite)).

    2. Relevant equations

    3. The attempt at a solution

    |f(x)-f(xo)| = |sin(1/x)- sin(1/xo)|= |2sin((xo-x)\2)cos((xo+x)/2)|

    =< 2|sin((xo-x)/(2xox))|=< |(xo-x)/(xox)|. is this inequality true? a similiar one is used in a different example. if it is, why? is it because sin(x)=<x ? when x is positive? now since x<1 choosing [tex]\delta[/tex]=xo[tex]\epsilon[/tex] then if

    |xo-x|<[tex]\delta[/tex] then |f(xo)-f(x)|<[tex]\epsilon[/tex]

    is this answer correct? what about the endpoints?
  2. jcsd
  3. Dec 20, 2007 #2
    i guess this is true for all xo and x different from 0 so it is true for (0, infinite) right?
  4. Dec 20, 2007 #3
    A cheap way would be to use the theorem that differentiability implies continuity
    The derivative of [tex]\sin(\frac{1}{x})[/tex] is [tex]\frac{-1}{x^2}\cos(\frac{1}{x})[/tex].
    The only place where the derivative is undefined is at x=0 which is not in the given interval so the function is continuous on the interval.

    I don't really like that way because it seems out of order (differentiability comes after continuity)

    Here's another.
    Definition of Continuity: The function is defined at every point on the interval and
    [tex]\lim_{x\to c}f(x) = f(c)[/tex]

    We know that the function is defined at every point except 0, so it is defined in the interval.
    The limit as x goes to c of the function is [tex]\sin(\frac{1}{c})[/tex] which is defined and equal to f(c) at every point except c=0, which isn't in the interval
    Therefore, the function is continuous on (0,1)
  5. Dec 20, 2007 #4
    sinx is continuous, and 1/x is continuous on (0,1). The composition of continuous functions is continuous. So sin(1/x) is continouous on (0,1)

    Are you required to prove that sinx itself is continuous?
  6. Dec 20, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    You say "(i know it's also continous on (0, infinite))". If a function is continuous on a set, A, it is continuous on any subset of A.
  7. Dec 20, 2007 #6
    ANY subset of A? What if the subset has an empty interior? How can f inverse of an open subset of f(A) (in the subspace topology of R) be open in A?
    Last edited: Dec 20, 2007
  8. Dec 20, 2007 #7
    What's the epsilon-delta proof for the continuity of sinx itself?
  9. Dec 20, 2007 #8


    User Avatar
    Science Advisor
    Homework Helper

    One can prove that |sin(x) - sin(y)| <= |x - y|.
  10. Dec 20, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    The empty set and A itself are still open subsets of A.
  11. Dec 20, 2007 #10
    Ok, I got it. If A is given the subspace topology of R, then any continuous function restricted to A is still continuous.

    But an epsilon-delta proof of this fact will not work if A is a set with an empty interior.
    Last edited: Dec 20, 2007
  12. Dec 25, 2007 #11
    i see you have all posted great replies, but can anyone tell me if my proof actually works. i want to understand the proper use of the epsilon-delta proof. the teacher will surely ask us to do such a proof. it seems okay.
  13. Dec 25, 2007 #12
    There's no need to use the epsilon-delta proof, and I don't think your prof expects it.
  14. Dec 25, 2007 #13
    I don't think so either, the exercise does not seem to encourage you to take a look at the epsilon-delta kind of proofs.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?