Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Hmm...

    Well, we obviously must find whether or not
    [tex]\lim_{x\rightarrow0}\sin{x}\sin{\frac{1}{x}}=0[/tex]

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Continuous Function
  1. Continuous functions (Replies: 7)

  2. Continuous function? (Replies: 0)

Loading...