Complex Indeterminate Algebraic Limit

[Liquid7800]

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:




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





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




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




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.

[Tom Mattson]

This limit comes out very straightforwardly with L'Hopital's Rule. Have you learned that yet?

[Liquid7800]

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!)

[Tom Mattson]

(I'm sorry...I am embarrassed to say I dont even know what a derivative is yet!)

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.

[Liquid7800]

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!

[Tom Mattson]

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

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

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

\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}

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 a^3-b^3 and a^4-b^4.

[Liquid7800]

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.

[Tom Mattson]

now "adding and subtracting one" is that a technique to remove radicals and turn it into a binomial for expansion?


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.


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.

It's got no name that I know of.