Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

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

    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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    1. For your numerator:
    Set:
    [tex]a=\sqrt[4]{n+2}[/tex]
    [tex]b=\sqrt[4]{n+1}[/tex]
    Show that (for example by polynomial division):
    [tex]a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})[/tex]
    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:

    [tex]
    \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)
    [/tex]

    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

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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:
    [tex]\frac{(n+3)^{\frac{2}{3}}+(n+3)^{\frac{1}{3}}n^{\frac{1}{3}}+n^{\frac{2}{3}}}{3((n+2)^{\frac{3}{4}}+(n+2)^{\frac{2}{4}}(n+1)^{\frac{1}{4}}+(n+2)^{\frac{1}{4}}(n+1)^{\frac{2}{4}}+(n+1)^{\frac{3}{4}})}[/tex]
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit with cube and fourth root
Loading...