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Continuity question

  1. Jan 13, 2012 #1
    my book says, the function y = x*sin(1/x) is not continuous at x = 0, however by defining a new function by



    F(x) =
    x*sin(1/x) , x ≠ 0
    0 , x = 0

    then F is continuous at x = 0.

    This does not make sense to me because the limit as x → 0 is equal to 1, not zero, so therefore there would be a jump discountinuity at x = 0.

    unless my calculus is wrong or my understanding is wrong... but isn't
    Lim [x*sin(1/x)] = Lim [(x/x)*sin(1/x)/(1/x)] = (1)*(1) = 1 as x approaches zero.
    x → 0
     
  2. jcsd
  3. Jan 13, 2012 #2

    CompuChip

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    I think your calculus is wrong, unfortunately

    For example, using the squeeze theorem: [itex]-x \le x \sin(1/x) \le x[/itex], therefore
    [tex]\lim_{x \to 0} -x \le \lim_{x \to 0} \le x \sin(1/x) \le \lim_{x \to 0} x[/tex]
    hence
    [tex]0 \le \lim_{x \to 0} \le x \sin(1/x) \le 0 \implies \lim_{x \to 0} x \sin(1/x) = 0[/tex]

    You are probably confused with
    [tex]\lim_{x \to 0} \frac{1}{x} \sin(x) = 1[/tex]
    but if you replace x -> 1/x what you have is the "opposite":
    [tex]\lim_{x \to \infty} \frac{1}{x} \sin(x) = 1[/tex]
     
    Last edited: Jan 13, 2012
  4. Jan 13, 2012 #3
    Im not sure what Im reading towards the end...

    I think it says,

    Lim as x approaches 0 of x*sin(1/x),
    If we let u = 1/x then as x approaches zero, u approaches infinity, so

    Lim as u appraoches infinity of sin(u)/u is eqaul to zero... is this correct?
     
  5. Jan 13, 2012 #4

    CompuChip

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    Sorry about that, I made a typo.
    Of course I meant
    [tex]\lim_{x \to 0} \frac{1}{x} \sin x = 1[/tex]
    instead of 0 (corrected in the above post now).

    And yes, I switched to u = 1/x which means [itex]x \to 0[/itex] changes to [itex]u \to \infty[/itex] (and you replace x with 1/u everywhere).
     
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