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!

Complex Indeterminate Algebraic Limit

  1. Feb 20, 2009 #1
    Hello,
    I am working on solving Limits algebraically (we haven't broken into derivatives yet or L'Hopital) and I encountered a problem that seemed very difficult to find similar type problems on the NET or in books...and I have many books...seems radicals and various odd even index types aren't that common.
    Anyway here is the problem:

    [tex]


    \lim_{x\rightarrow 0} \frac{\sqrt[3]{1+x^2}- \sqrt[4]{1-2x}}{x+x^2}


    [/tex]


    2. In speaking with my teacher, she mentioned I couldn't use a conjugate because there is a mixture of odd and even radical indexes...nor could I use substitution because the terms under the radical are different so those attempts to get rid of the radical didn't work for me.

    My instructor mentioned if I split the problem into a form like

    [tex]


    \frac{(a^3 -1) - (a^4-1)}{x+x^2}


    [/tex]

    a^3 -1 and a^4 -1 for the numerator...then that might be a way.


    3. So my two questions are:

    What does she mean by splitting the problem up like this and is there a more creative way to take care of this radical? I am very curious for creative/ new ways to approach this problem!

    Thanks for any and all help.
     
  2. jcsd
  3. Feb 20, 2009 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This limit comes out very straightforwardly with L'Hopital's Rule. Have you learned that yet?
     
  4. Feb 20, 2009 #3
    No...unfortunately we are not there yet...and you are correct, l'hoptial's rule does figure it out ive heard for limits...but we just start derivitives this week...and I think l'hopital uses derivatives? (I'm sorry...I am embarrassed to say I dont even know what a derivative is yet!)
     
  5. Feb 20, 2009 #4

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm embarrassed to say I didn't notice that you said that in your first post. I'm further embarrassed to say that I can't think of another way to do this limit. :blushing:

    Is this a homework problem that's due soon? If not, then how about putting it on the back burner until you learn L'Hopital. It really does work out very easily that way.
     
  6. Feb 20, 2009 #5
    lol, thanks for taking a look at this. anyway.

    It is a homework problem (due mon.) so I will tackle it regardless.
    Is L'Hopital hard to learn if you havent gone through the derivatives section yet?

    Now I found some good examples on this site that do something to solve similar problems:

    http://www.pinkmonkey.com/studyguides/subjects/calc/chap2/c0202603.asp

    That show complex and similar problems...but they dont go into detail on HOW they solve it and they show the " ' " symbol which I think is a derivative? Is this L'Hopital they are using here?

    I appreciate you looking at this anyway!
     
  7. Feb 20, 2009 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Oh! The method of Example 7 looks promising. What you need to do is add and subtract 1 from the numerator as follows.

    [tex]\lim_{x\rightarrow 0} \frac{\left(\sqrt[3]{1+x^2}-1\right)-\left(\sqrt[4]{1-2x}-1\right)}{x+x^2}[/tex]

    Then I would split this up into 2 limits and try to do them separately.

    [tex]\lim_{x\rightarrow 0} \frac{\left(\sqrt[3]{1+x^2}-1\right)}{x+x^2}-\lim_{x\rightarrow 0} \frac{\left(\sqrt[4]{1-2x}-1\right)}{x+x^2}[/tex]

    Now apply the method of Example 7. I did it and got the correct answer, which is 1/2 (Thanks, Maple). You'll have to recall how to factor expressions of the form [itex]a^3-b^3[/itex] and [itex]a^4-b^4[/itex].
     
  8. Feb 20, 2009 #7
    Hey thanks I appreciate the help...that does look like a good idea and it corresponds to what my teacher said too..
    now "adding and subtracting one" is that a technique to remove radicals and turn it into a binomial for expansion? Does this technique have a name? Or is it something that logically makes sense and needs no name because it wouldnt be possible if there were a "+" inbetween the numerator.
     
  9. Feb 20, 2009 #8

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, the point is to try to get something on which you can use the factoring formulas for the difference of 2 cubes and difference of 2 squares.

    It's got no name that I know of.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex Indeterminate Algebraic Limit
Loading...