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Limits of a sin function

  1. Sep 17, 2006 #1
    Consider the function f(x) = sin(1/x).

    (a) Find a sequence of x-values that approach 0 such that sin (1/x) = 0.
    [Hint: Use the fact that sin(pi) = sin(2pi) = sin(3pi) = ... = sin (npi) = 0.]

    (b) Find a sequence of x-values that approach 0 such that sin(1/x) = 1.
    [Hint: Use the fact that sin (npi/2) = 1 if n = 1,5,9....]

    (c) Find a sequence of x-values that approach 0 such that sin(1/x) = -1.

    (d) Explain why your answers show that the limit as x->0 does not exist.


    Here are my thoughts.

    (a) it tells me that sin(pi) = sin(2pi) = sin(3pi) = ... = sin(npi) = 0.
    but, my function is sin(1/x), is there any way I can figure out what the values are for this function using the values they already gave me for sin(x)?.. I'm guessing that's what they want me to do, I'm just not sure how..

    (b) same thoughts as (a)..

    (c) dunno :(

    (d) cant get this one if i don't have the other ones first..

    I'm totally confused.. *sigh*
    any help appreciated, thanks.
     
  2. jcsd
  3. Sep 18, 2006 #2

    Galileo

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    Using the hint, you can see that you can use the first x in your sequence such that 1/x=pi. For then: sin(1/x)=sin(pi)=0, see? The second x: 1/x=2pi etc.
    So figure out what the sequence should be x_n.
    You can use the same reasoning with the other questions.
     
  4. Sep 18, 2006 #3
    all right i continued work on this problem... here's what it's looking like now.

    (a) I set 1/x = pi, 2pi, 3pi, npi
    1/x = pi ---> x = 1/pi
    1/x = 2pi ---> x = 1/(2pi)
    1/x = 3pi ---> x = 1/(3pi)
    1/x = npi ---> x = 1/(npi)

    Answer: x = 1/pi, 1/(2pi), 1/(3pi)... 1/(npi).

    (b) I set 1/x = pi/2, (5pi)/2, (9pi)/2, (npi)/2
    1/x = pi/2 ---> x = 2/pi
    1/x = (5pi)/2 ---> x = 2/(5pi)
    1/x = (9pi)/2 ---> x = 2/(9pi)
    1/x = (npi)/2 ---> x = 2/(npi)

    Answer: x = 2/pi, 2/(5pi), 2/(9pi), 2/(npi).

    (c) First, I set sin(x) = -1 to find some values.
    sin x = -1
    x = (3pi)/2, (7pi)/2, (11pi)/2
    So, my hint here is sin((npi)/2) = -1.. if n = 3, 7, 11...
    1/x = (3pi)/2 ---> x = 2/(3pi)
    1/x = (7pi)/2 ---> x = 2/(7pi)
    1/x = (11pi)/2 --> x = 2/(11pi)
    1/x = (npi)/2 ---> x = 2/(npi)

    Answer: x = 2/(3pi), 2/(7pi), 2/(11pi), 2/(npi)

    (d) Still not sure about this one...
     
  5. Sep 18, 2006 #4

    Galileo

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    Here, the values n takes are not 1,2,3,.. So I'd rewrite it the denominator in 2/(npi) so it takes the values 2/pi, 2/(5pi), 2/(9pi)..., for n=1,2,3,... resp.
    Same comment holds for c).

    What does it mean for a limit to exist? In this case, what does it mean when I say [itex]\lim \limits_{x\to 0}\sin(1/x)=L[/itex], for some number L?
     
  6. Sep 18, 2006 #5
    I'm not really getting what you're saying in the first part... Is my answer correct, just not very elaborate? Are you just trying to say that I should note somewhere that n is not going 1,2,3, and it's going 1, 5, 9, n?
    or am i missing the point?

    And, as for the 2nd comment.. I think I know the answer.. how does this sound?
    Each part of the problem asks as x approaches 0, y is a different value (ie, (a) = 0, (b) = 1, (c) = -1). If the value of sin(1/x) as x approaches 0 is 3 different numbers (0,1,-1), the limit cannot exist.
     
  7. Sep 18, 2006 #6

    Galileo

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    A sequence is usually denoted like (for example the first one) [tex]x_n=\frac{1}{n\pi}[/tex], and it's implicit that n takes the values 1,2,3,..
    So if you write x_n=2/(npi) for (b), it's wrong unless you add that n takes the values 1,5,9 etc, but it's nicer to change your expression for x_n so that [tex]x_1=2/\pi, x_2=2/(5\pi), x_3= 2/(9pi) [/tex]
    Well, that's true, but you kinda jump right to the conclusion from what is given (even though it's a small jump). Can you elaborate a bit on the reasoning? What is it in the definition of the limit that shows that the limit in this example doesn't exist?
     
    Last edited: Sep 18, 2006
  8. Sep 18, 2006 #7
    so, for the explanation...
    There are x-values approaching 0 where sin(1/x) = -1. There are also x-values approaching 0 where sin(1/x) = 1. There are also x-values approaching 0 where sin(1/x) = 0. So, if the limit existed, it would have to be -1, 1, and 0. A function cannot approach 3 different numbers as x approaches one number. So, the limit does not exist.
     
  9. Sep 18, 2006 #8

    Galileo

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    If you've never had a precise definition of a limit, then I guess that's the best answer one can give.

    The definition of [tex]\lim \limits_{x\to 0}\sin(1/x)=L[/tex], says that we can make sin(1/x) as close to L as we want for all x in the interval (-e,e) for some number e>0. See how we can use this to show the given limit doesn't exist?
     
  10. Sep 18, 2006 #9
    actually, we did do the whole episilon-delta definition of a limit.. and i didn't quite grasp it fully.. but, in general, i know it said something like.. if you give me any distance |f(x) - L|.. i can give you a horizontal distance |x-c|.
     
  11. Sep 19, 2006 #10

    Galileo

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    Then I would suggest to really study this definition, because limits are of fundamental importance in calculus.

    In general: [tex]\lim \limits_{x\to a}f(x)=L[/tex] says that we can make
    f(x) as close to L as we like if x is close enough (but not equal) to a.

    In other words: We can make |f(x)-L| as small as we like (smaller than any given positive number [itex]\delta[/itex]) for all x in an open interval around a: x in (a-e,a+e), except maybe x=a (note that this last thing is the same as saying 0<|x-a|<e)

    So in mathematical terms:
    For any given [itex]\delta>0[/itex], there exists an [itex]\epsilon>0[/itex] such that [tex]0<|x-a|<\epsilon \Rightarrow |f(x)-L|<\delta[/tex]

    Check your textbook for precise details concerning the domain of f. In any case, the reason it goes wrong for sin(1/x) is that for a given delta we must have |sin(1/x)-L|<delta for ALL x satisfying 0<|x|<e. Using the sequences you got, can you show such an epsilon does not exist?
     
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