1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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




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

  2. Continuous function? (Replies: 0)

Loading...