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Continuity - analysis

  1. Jul 2, 2008 #1
    How come sin(x^-1) is not continuous and xsin(x^-1) is?
     
  2. jcsd
  3. Jul 2, 2008 #2

    matt grime

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    Calling something a function, and saying it is continuous, all require you to specify what the domains are that you're talking about. Any domain that I can reasonably use to make sin(1/x) a function mean it is continuous.

    I suspect that you want to define it as sin(1/x) for x=/=0 and 0 when x=0, as a function from R to R.

    It isn't then continuous because it fails to satisfy the definition of continuous.

    If you want to prove this, then use the sequential definition of limit - it should be an easy and illuminating exercise.
     
  4. Jul 3, 2008 #3
    In the sequential definition of a limit, n approaches infinity and we show there exists a natural number, say p, such that n>=p implies |f(n)-L| < epsilon

    Do you mean to prove sin(x^-1) is not continuous we show we cannot find a real number r, such that x<=r implies
    |sin(x^-1) -0|< epsilon ?
     
  5. Jul 3, 2008 #4

    HallsofIvy

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    That is NOT a definition of "continuous". For one thing a function is defined to be continuous at specific points. That would be, if n is not constrained to be an integer, the definition of "limit as x goes to infinity". Since functions are not defined "at infinity" what you give is not even a definition of "continuous at infinity" (since no such thing exists).

    A function is continuous at x= a if [itex]\lim_{x\rightarrow a} f(x)[/itex], f(a) exists, and [itex]\lim_{x\rightarrow a} f(x)= f(a)[/itex]. A function is not continuous at x= a if any one of those is not true. A function is said to be "continuous" if it is continuous at every value of x.

    At what value of x is sin(x-1) not continuous?
     
  6. Jul 3, 2008 #5
    Yeah. I think when proving continuity its similar to the proof of a limit. In the case of sin (x^-1) our epsilon will always be smaller than our delta rather than the other way around which is what is required to prove the existence of a limit/continuity. Is this correct !?
     
  7. Jul 4, 2008 #6

    HallsofIvy

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    ?? There is nothing said about "epsilon" or "delta" being smaller in the definition of a limit. We only require that "given epsilon" there exist delta with such that if
    |x- a|< delta, then |f(x)-f(a)|< epsilon. You need to go back and check the definitions of both "limit" and "continuous".
     
  8. Jul 4, 2008 #7
    so in the case of sin(x^-1)
    |x| < delta must imply |sin(x^-1)| < epsilon which is impossible because..........?
     
  9. Jul 4, 2008 #8

    HallsofIvy

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    First, as has been said before, just "sin(x-1)" is not continuous at x= 0 because it is not DEFINED there. The more important point is that that is not a "removable discontinuity"- it cannot be defined at x= 0 in any way that will make it continuous. And that is because no matter how small delta is, between 0 and delta, sin(1/x) will take on all the values sin(y) does for y between 1/delta and infinity. (You have 0< x< delta so 1/delta< 1/x< infinity and you can take y= 1/x.)

    At some points larger than 1/delta, you will have values of y such that sin(y)= 1 and some at which sin(y)= -1. That means that for 0< x< delta, you will have some values of x so that sin(1/x)= 1 and some for which sin(1/x)= -1. As soon as you take epsilon< 1/2, you can't have |sin(1/x)-L|< epsilon for any delta. If |1- L|< 1/2, |-1- L|= |1+ L| can't be less than 1/2.

    Now, f(x)= xsin(1/x) is also no continuous at x= 0 because it is not defined there. However, that is now a "removable discontinuity". While sin(1/x) takes on all possible values between -1 and 1 because it does stay between them, we now have -x<= x sin(1/x)<= x. And x itself goes to 0. [itex]\lim_{x\rightarrow 0} x sin(1/x)= \lim_{x\rightarrow 0} x= 0. Defining f(0)= 0, f(x)= x sin(1/x) if x is not 0, makes it a continuous function at 0 and so for all x.
     
  10. Jul 4, 2008 #9
    Yeah like theres no value of x (say p) such that

    |x|<p implies |sin(x^-1)|<1/2
     
  11. Jul 4, 2008 #10
    sin(x^-1) and xsin(x^-1)

    Let us look at the graphs of these functions.

    sin(x^-1) OSCILLATES between -1 and +1 on the y axis
    it takes the value 1 at x = 0.67, 0.22, 0.133...
    Therefore if we have a restriction like you mentioned say epsilon < 1/2 there is no RANGE of values of x < p as every so often the corresponding y value (sin(x^-1)) will jump to 1

    However looking at xsin(x^-1) this is like the standard sine graph DAMPED approaching zero. In this case we can set up a range of values for delta, limited by a restriction on epsilon
     
  12. Oct 5, 2010 #11
    so in short, sin(1/x) is NOT continuous even if it is defined by f(x)=0 when x=0? and xsin(1/x) IS continuous if defined by f(x)=0 when x=0? am I right or wrong...sorry I have the same problem understanding this
     
  13. Nov 15, 2010 #12
    sin(1/x) is no continous because of you let x_n = 1/2npi

    sin(1/x_n) =0

    If x_n = 1/ (pi/2 +2npi)

    sin(1/x_n) = 1.
    A slightly naive explaination of this is that...
    Basically your function has two different subsequencial limits although x_n -> 0. So depending on the way zero is approached you have different limits which means your function cannot possible be continous.
     
  14. Nov 16, 2010 #13

    HallsofIvy

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    In case the point hasn't been made clearly, NEITHER of these functions is continuous at x= 0. x sin(x^-1) has a removable discontinuity at x= 0. The function defined by "f(x)= x sin(x^-1) if x is not 0: and "f(0)= 0" is continous for all x but that is no longer the same function.
     
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