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Homework Help: Not so easy limit to inf

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}-7^{3n}+(n+1)!\right) \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n [/tex]

    2. Relevant equations
    Limits manipulation

    3. The attempt at a solution

    Ok, in the first parenthesis, I have clear (I hope so) that the biggest term is the one containing [itex]n^n[/itex], since [itex]n^n>n!>a^n>....[/itex]
    I should make passages, but this is quite clear to me.

    The hard part is that
    [tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n [/tex]

    I am tempted to replace [itex]x = \frac{1}{n}[/itex]
    so that I obtain

    [tex]\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} [/tex]

    Now, remembering that for [itex]x \to 0[/itex]
    [tex]\sin x = x - \frac{x^3}{6}+ o(x^3)[/tex]
    I'll write
    [tex]\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} = \left(\frac{x^3}{6}+ o(x^3) \right)^{\frac{1}{x}} [/tex]

    Let me forget the rest [itex]o(x^3)[/itex], and go back to [itex]n[/itex]
    and write that

    [tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n = \frac{1}{6n^{3n}}[/tex]
    This should hold true, as [itex]n \to +\infty[/itex]

    Now, I go back to the first part of the limit

    [tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}-7^{3n}+(n+1)!\right) [/tex]
    neglecting the parts that are lower oder of infinity wrt to [itex]n^n[/itex]I can write it as
    [tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}\right) [/tex]
    [tex]\lim_{n \to +\infty} \left (7(6^{n})n^{3n}\right) [/tex]

    If I divide it by the term coming from the other expression I get

    [tex]\lim_{n \to +\infty} \frac {\left (7(6^{n})(n^{3n})\right)}{6n^{3n}} = \lim_{n \to +\infty} \left (\frac{7}{6}(6^{n})\right) = +\infty [/tex]

    I should have finally found that all that stuff goes to infinity.

    But I'm not really sure of the passages.... anyone can kindly confirm ?
    Last edited: Aug 15, 2011
  2. jcsd
  3. Aug 15, 2011 #2


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    Homework Helper

    I believe this is [tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n = \lim_{n \to +\infty} \left(\frac{1}{6n^{3}}\right)^n = \lim_{n \to +\infty} \left(\frac{1}{6^{n}n^{3n}}\right) ,[/tex]

    making the final result

    [tex]\lim_{n \to +\infty} \frac {\left (7(6^{n})(n^{3n})\right)}{6^{n}n^{3n}} = 7 .[/tex]

    (It took a while to spot the omission. A graph of the function confirms this limit -- although the grapher's calculator gave out before x = 60...)
  4. Aug 15, 2011 #3
    Ah sure.... poor me !

    Thanks so much.
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