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Lim x->+inf xsin(1/x)

  1. Nov 25, 2008 #1
    Hello,

    [tex]\[
    \mathop {\lim }\limits_{x \to + \infty } x\sin ({\textstyle{1 \over x}})
    \]
    [/tex]

    In my textbook they show a way to do this limit using variable substitution. I understand how they did it,

    however, before I saw what they did, I tryed to work it out intuitively.... as x goes to infinity, 1/x goes to 0... since sin(0) = 0, then the limit must be equal to 0

    the correct answer is 1, but I don't understand why the intuitive method fails. can somebody help me out?

    thank you, sorry for bad english
     
  2. jcsd
  3. Nov 25, 2008 #2

    HallsofIvy

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    "Intuitively" sin(1/x) goes to 0. It does not follow that x sin(1/x) goes to 0 because x is going to infinity. You can't say "[itex]\infty*0= 0[/itex]".

    Your English is excellent.
     
  4. Nov 25, 2008 #3
    Apply the La'Hospital rule

    for that first of all convert the equation to form such that after applying limit directly we get 0/0 or infinity/infinity form. Then differentiate both the numerator and the denomenator and then apply the limit thus

    f(x) = xsin(1/x) convert to f(x)/g(x) form i.e.
    f(x)/g(x) = sin(1/x)/1/x which is now in the form of 0/0 ,then according to La'Hospital rule
    f(x)/g(x) = f'(x)/g'(x) thus
    f'(x)/g'(x) = [-1/x2cos(1/x)]/-1/x2

    this gives

    f'(x)/g'(x) = cos(1/x) now apply the limit to this derivative
    which gives cos(1/infinity) = cos(0) = 1 which is the answer
     
  5. Dec 2, 2010 #4
    Sorry to bump such an old thread, but is there a way to show that the function x*sin(1/x) tends to 1 as x goes to infinity without using l'Hospital's rule?
     
  6. Dec 2, 2010 #5

    micromass

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    Do you know the formula

    [tex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1[/tex]

    if you don't know it, then I don't think you can do the question...
     
  7. Dec 2, 2010 #6
    I remember it from high school, but we haven't proven it in class at the university. We have done the limit of this function as x approaches 0, however, and with that one I don't have a problem. Is there perhaps a way you'd be able to squeeze xsin(1/x) between two functions that converge to 1 as x approaches plus infinity?
     
  8. Dec 2, 2010 #7

    micromass

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    You could probably prove it with the following inequality

    [tex]|\sin(1/x)|\leq 1/|x|\leq |\tan(1/x)|[/tex]

    The proof of this is nontrivial... You can see it in the following great video:
     
    Last edited by a moderator: Sep 25, 2014
  9. May 17, 2012 #8
    squuezing theorem

    using squuezing theorem

    limx→∞ tan-1(x)
    x

    solve.
     
  10. May 17, 2012 #9

    Ray Vickson

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    For small |y|, [itex] \sin y = y + O(|y|^3),[/itex] so for large x, [itex] \sin(1/x) = 1/x + O(1/x^3).[/itex] That implies [itex] x \sin(1/x) = 1 + O(1/x^2) \rightarrow 1[/itex] as [itex] x \rightarrow \infty.[/itex]

    RGV
     
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