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The Topologist's Sine Curve

  1. Apr 5, 2013 #1
    On Page 106 in baby rudin (diff. chapter) when he tries to calculate the derivative of the fuction

    $$f(x) = \begin{cases}
    x^2 sin(\frac{1}{x}) & \textrm{ if }x ≠ 0 \\
    0 & \textrm{ if }x = 0 \\

    rudin used the absolute value in trying to compute the limit as ##t → 0##


    ##\left|\frac{f(t) - f(0)}{t - 0}\right| = \left|t \ sin(\frac{1}{x})\right| ≤ |t|##

    Why the abs. value?
  2. jcsd
  3. Apr 6, 2013 #2


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    I think it has to see with the fact that sin(-x)=-sin(x) . The limit then will be non-negative. Since is negative
    in the 4th quadrant.
  4. Apr 8, 2013 #3
    Absolute convergence implies convergence as in chapter 3 of the same book. And it turns out in this example checking absolute convergence is more obvious.
  5. Apr 9, 2013 #4


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    It just shows that the Newton quotient at zero is squeezed by a number that goes to zero with t.
  6. Apr 21, 2013 #5
    Thank you all
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