1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Natural log limits as n approaches infinity

  1. Oct 10, 2013 #1
    My question is:

    Show the limit of
    [tex]x_{n}=\frac{ln(1+\sqrt{n}+\sqrt[3]{n})}{ln(1+\sqrt[3]{n}+\sqrt[4]{n})}[/tex]
    as n approaches infinity


    Solution:
    [tex]{x_n} = \frac{{\ln (1 + {n^{\frac{1}{2}}} + {n^{\frac{1}{3}}})}}{{\ln (1 + {n^{\frac{1}{3}}} + {n^{\frac{1}{4}}})}} = \frac{{\ln \left( {{n^{\frac{1}{2}}} \cdot (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})} \right)}}{{\ln \left( {{n^{\frac{1}{3}}} \cdot (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{\frac{1}{{12}}}}}})} \right)}} = [/tex]


    [tex]= \frac{{\ln {n^{\frac{1}{2}}} + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\ln {n^{\frac{1}{3}}} + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} = \frac{{\frac{1}{2}\ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3}\ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} =[/tex]

    [tex]= \frac{{\frac{1}{2} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{2}}}}} + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{3}}}}} + \frac{1}{{{n^{\frac{1}{{12}}}}}})}}[/tex]


    Now I'm stuck.
    Is this the right way to do it?
     
  2. jcsd
  3. Oct 10, 2013 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That looks good so far.... as n goes to infinity, what happens to 1/ln(n), and what happens to ln(1+1/n1/2+1/n1/6) and the similar term in the denominator?
     
  4. Oct 10, 2013 #3

    Mark44

    Staff: Mentor

    How did you (OP) go from (1/2)ln(n) to 1/2 + 1/ln(n) ? And same question for the similar part of the denominator?
     
  5. Oct 10, 2013 #4

    Mark44

    Staff: Mentor

    I would continue from here:
    [tex]= \frac{{ \frac 1 2 \ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{ \frac 1 3 \ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} [/tex]
    by factoring ln(n) from each term in the numerator and each term in the denominator, with the result looking like this:
    $$ \frac{\frac 1 2 ln(n)(1 + \text{other stuff})}{\frac 1 3 ln(n)(1 + \text{some other stuff})}$$
    As long as you can convince yourself the <other stuff> goes to zero as n gets large, you should be able to get the desired answer of 3/2.
     
  6. Oct 10, 2013 #5

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Mark he just divided the numerator and denominator by ln(n)
     
  7. Oct 10, 2013 #6

    Mark44

    Staff: Mentor

    OK, I didn't see that. It looked to me like he was using an invalid log property. What he did is pretty close to what I'm suggesting in post #4.
     
  8. Oct 10, 2013 #7
    I liked your first 3 steps until you got to ##\frac{(1/2)logn + stuff}{ 1/3logn + stuff}##. What I'm calling "stuff" will all go to 0 as n ## \rightarrow \infty##.

    So from that point you can go directly to the answer of 3/2 -- I don't thing you need to fuss any further.

    You should probably explain why the stuff goes to 0.
     
  9. Oct 10, 2013 #8
    Many thanks to all of you for the quick response! I can see it now!

    These kind of problems are quite new for me so maybe I overdo it.

    I need to do some more of these so I can get some grip of it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Natural log limits as n approaches infinity
Loading...