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Sinx /x limit

  1. Feb 15, 2014 #1


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    Gold Member

    Hello, I am just having a small confusion in extraction of the limit:
    [itex] lim_{x→∞} \frac{sinx}{x}[/itex]

    One way to do it is by the sandwich rule:
    [itex] - \frac{1}{x} ≤ \frac{sinx}{x}≤ \frac{1}{x}[/itex]
    from where you get that by taking the limit on both sides:
    [itex] lim_{x→∞} \frac{sinx}{x}=0[/itex]

    Now on the other hand I'd like to write the sin through one of its definitions:
    [itex]sinx=∑_{n=0} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}[/itex]
    and deal it as a polynomial.... In that case, by taking the limit of x approaching infinity, I will get infinity instead, since every [itex]x^{k}, k>1[/itex] will cancel out the denominator...
    In fact it will be like:
    [itex]lim_{x→∞} \frac{1}{x} ∑_{n=0} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}[/itex]
    [itex]lim_{x→∞} \frac{x-x^{3}/6 + ...}{x}[/itex]
    [itex]lim_{x→∞} (1-x^{2}/6 + ...)[/itex]
    which in fact is undefined...

    What is the problem of the 2nd approach?
    Last edited: Feb 15, 2014
  2. jcsd
  3. Feb 15, 2014 #2


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    Homework Helper

    The problem with the second approach is that if [itex]x[/itex] is sufficiently large then
    \frac{x^n}{n!} < \frac{x^{n+2}}{(n+2)!}
    so you can't ignore the tail.

    EDIT: more precisely,
    \lim_{x \to \infty} \lim_{N \to \infty} \sum_{n=0}^N \frac{x^{2n}}{(2n+1)!} = 0
    \lim_{N \to \infty} \lim_{x \to \infty} \sum_{n=0}^N \frac{x^{2n}}{(2n+1)!}
    does not exist.
    Last edited: Feb 15, 2014
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