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Limit with cube and fourth root

  1. Nov 18, 2004 #1
    Hi guys, I have another limit I can't move with. Well, I guess it goes to zero, but can't show a bulletproof evidence:

    \lim_{n \rightarrow \infty} \frac{ \sqrt[4]{n + 2} - \sqrt[4]{n + 1}}{ \sqrt[3]{n + 3} - \sqrt[3]{n}}

    Even after I got rid of denominator, I can't find some known lemma to show that this limit is really 0. Will somebody help me to find any?

    Thank you.
  2. jcsd
  3. Nov 18, 2004 #2


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    1. For your numerator:
    Show that (for example by polynomial division):
    2. Use a similar technique for your denominator.
    It should now be quite simple to evaluate your limit.
  4. Nov 18, 2004 #3
    Yes I used this method to get rid of denominator, hence I got this:

    \frac{1}{3} \lim_{n \rightarrow \infty} \left( \sqrt[4]{n+2}.\sqrt[3]{(n+3)^2} + \sqrt[4]{n+2}.\sqrt[3]{n^2+3n} + \sqrt[4]{n+2}.\sqrt[3]{n^2} - \sqrt[4]{n+1}.\sqrt[3]{(n+3)^2} - \sqrt[4]{n+1}.\sqrt[3]{n^2+3n} - \sqrt[4]{n+1}.\sqrt[3]{n^2} \right)

    I tried to divide it somehow with the largest term but the only thing I got were undeterminate forms like [itex]0. \infty[/itex]
  5. Nov 18, 2004 #4
    please....... i dont know how to solve the limits that its denominator is from 8th or 5th or...etc degree!!!! what should i do to solve such a quesion?!!!
  6. Nov 18, 2004 #5


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    Let's take a second look at that limit:

    [tex]\lim_{n\rightarrow\infty} ((n+2)^{\frac{1}{4}}-(n+1)^{\frac{1}{4}})\frac{1}{3}((n+3)^{\frac{2}{3}}+(n)^{\frac{1}{3}}(n+3)^{\frac{1}{3}}+n^{\frac{2}{3}})[/tex]

    Can be bounded above since [itex](n+3)^\frac{1}{3}>(n)^\frac{1}{3}[/itex]
    [tex] \lim_{n\rightarrow\infty} ((n+2)^{\frac{1}{4}}-(n+1)^{\frac{1}{4}})(n+3)^{\frac{2}{3}}[/tex]
    But that's equal to
    [tex]\lim_{n\rightarrow\infty} \frac{(n+3)^{\frac{2}{3}}}{((n+2)^\frac{1}{4}+(n+1)^\frac{1}{4})((n+2)^\frac{1}{2}+(n+1)^\frac{1}{2})}[/tex]
    which is bounded above by
    [tex]\lim_{n \rightarrow\infty} \frac{(n+3)^{\frac{2}{3}}}{4(n+1)^\frac{3}{4}}[/tex]
    Which has a larger exponet on bottom, so it goes to zero.
  7. Nov 19, 2004 #6


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    But this is not at all what you should do!!
    You should get rid of the difference both in the denominator AND numerator.
    Then you'll end up with:
    This is easy to evaluate, the leading order behavior is [tex]\frac{1}{4n^{\frac{1}{12}}}, n\to\infty[/tex]
    Last edited: Nov 19, 2004
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