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(x_n)->0 and lim(x_n)sin(1/(x_n))=0 help?

  1. Oct 12, 2008 #1
    (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    1. The problem statement, all variables and given/known data
    Let [tex]x_n[/tex] be a sequence in R with [tex]x_n \rightarrow 0 [/tex] and [tex] x_n \neq 0[/tex] for all n. Prove that [tex] lim (x_n) sin \frac{1}{x_n} = 0[/tex].

    3. The attempt at a solution

    I think I might have a solution if I say that if [tex]x_n \rightarrow 0[/tex], then [tex] \frac{1}{x_n} \rightarrow \infty[/tex].

    Then [tex] sin \frac{1}{x_n} \times (x_n) \rightarrow 0 [/tex]. This might or might not be a good method.

    Anyway, ... I am required to use the squeeze thrm to prove this. Could somebody give me a hint on which two outer limits to use?
     
  2. jcsd
  3. Oct 12, 2008 #2
    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    Are you sure that xn converges to 0? Suppose the sequence is 1, 1/2, 1/3, ... which converges to 0. The sine sequence is sin 1, sin 2, sin 3, ... which I highly doubt converges to 0.
     
  4. Oct 12, 2008 #3
    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    The problem states that "if" x_n approaches 0 then the limit of (x_n)sin(1/x_n) approaches 0. I assume we have to use that information.
     
  5. Oct 12, 2008 #4

    gabbagabbahey

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    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    Hint: try to show that [itex]-abs(x_n) \leq x_nsin \left(\frac{1}{x_n} \right) \leq abs(x_n)[/itex] for all n. If you can do that, then just apply the squeeze theorem and your done.

    Edit- you can't just say that lim x_n ->0 of x_nsin(1/x_n) is zero because x_n goes to zero....the limit of sin(1/x_n) does not exist so [itex]\lim ( x_n sin(1/x_n)) \neq (\lim (x_n))(\lim (sin(1/x_n))) [/itex]
     
  6. Oct 12, 2008 #5
    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    Please let me know if I need more rigor on this...

    "Proof"

    Since [tex]sin\frac{1}{x_n}[/tex] is always between -1 and 1,

    [tex]x_nsin\frac{1}{x_n} \leq x_n[/tex]

    Using similar logic,

    [tex]x_n \leq x_nsin\frac{1}{x_n} \leq x_n[/tex]

    Since both negative and positive [tex]x_n \rightarrow 0[/tex]

    we know by the squeeze theorem that

    [tex]x_nsin\frac{1}{x_n} \rightarrow 0[/tex]

    EDIT: How do I incorporate the abs?
     
  7. Oct 12, 2008 #6

    gabbagabbahey

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    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    abs means absolute value (x_n could be negative for some n, so you must take the absolute value...suppose that there were a value of n for which x_n=-2 and sin(1/x_n)=+0.5, clearly x_n sin(1/x_n)=-1 which is not less that x_n=-2, but it is less than |-2|=2 )

    [tex]\Rightarrow -|x_n| \leq x_nsin\frac{1}{x_n} \leq |x_n|[/tex]

    not

    [tex]-x_n \leq x_nsin\frac{1}{x_n} \leq x_n[/tex]

    If x_n ->0, then so do Abs(x_n) and -Abs(x_n)
     
    Last edited: Oct 12, 2008
  8. Oct 14, 2008 #7
    Re: (x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

    Okay, so here is what I have...

    Proof

    Since

    [tex]\sin\frac{1}{x_n}[/tex]

    is always between -1 and 1,

    [tex]x_n\sin\frac{1}{x_n} \leq |x_n|[/tex]

    Using similar logic,

    [tex]-|x_n| \leq x_n\sin\frac{1}{x_n} \leq |x_n|[/tex]

    Since

    [tex]-|x_n| \to 0[/tex]

    and

    [tex]|x_n| \to 0[/tex]

    , we know by the squeeze theorem that

    [tex]x_n\sin\frac{1}{x_n} \rightarrow 0[/tex]
     
    Last edited: Oct 14, 2008
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