1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

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


    User Avatar
    Science Advisor
    Homework Helper

    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]
    [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


    User Avatar
    Science Advisor
    Homework Helper

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

Similar Threads - Continuity question Date
Uniform Continuity question May 2, 2017
Limit and continuity question Mar 14, 2015
Analysis Question on Continuity Nov 7, 2014
X^2 is continuous question Nov 25, 2013
Question regarding the continuity of functions Sep 10, 2013