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Continuous Function

  1. Feb 27, 2005 #1
    Using the inequalities -1 = < sin 1/x = < 1 for x not equal to 0 determine wether or not the function
    f(x)= {sinxsin1/x x is not equal to 0
    f(x)= {0 x=0

    continuous at x=0
  2. jcsd
  3. Feb 28, 2005 #2

    Well, we obviously must find whether or not

    Which is a bit of a challenge since [tex]\lim_{x\rightarrow\infty}\sin{\frac{1}{x}}[/tex] is undefined.
    Looks like squeeze theorem time.
    For all x, [tex]-|x|\leq\sin{x}\leq|x|[/tex] and [tex]|\sin{\frac{1}{x}}|\leq1[/tex].
    Therefore, [tex]-|x|\leq\sin{x}\sin{\frac{1}{x}}\leq|x|[/tex] for all x.
    Since [tex]\lim_{x\rightarrow0}-|x|=\lim_{x\rightarrow0}|x|=0, \lim_{x\rightarrow0}\sin{x}\sin{\frac{1}{x}}=0[/tex]

    And thus it is continuous
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