# Complex Indeterminate Algebraic Limit

• Liquid7800
In summary, a student is working on solving limits algebraically and is having trouble finding similar problems online or in books. The problem involves a mixture of odd and even radical indexes and the student's teacher suggested splitting the problem into a specific form. The student is curious about alternative methods for solving the problem and someone suggests using L'Hopital's Rule, which the student has not yet learned. The student then comes across an example that uses a technique of adding and subtracting 1 to the numerator in order to factor and solve the limit. The student is grateful for the help and asks for clarification on the technique. The technique does not have a specific name, but is a way to simplify and solve certain limits.

#### 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.

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

No...unfortunately we are not there yet...and you are correct, l'hoptial's rule does figure it out I've 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 don't even know what a derivative is yet!)

Liquid7800 said:
(I'm sorry...I am embarrassed to say I don't 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.

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.

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 haven't 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 don't 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!

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 $a^4-b^4$.

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 wouldn't be possible if there were a "+" inbetween the numerator.

Liquid7800 said:
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 wouldn't be possible if there were a "+" inbetween the numerator.

It's got no name that I know of.

## What is a complex indeterminate algebraic limit?

A complex indeterminate algebraic limit is a type of limit where the variable in the expression approaches a complex number, and the limit itself cannot be determined by simple algebraic manipulations.

## How is a complex indeterminate algebraic limit different from a regular limit?

In a regular limit, the variable approaches a real number, and the limit can be determined by simple algebraic manipulations. In a complex indeterminate algebraic limit, the variable approaches a complex number, and the limit cannot be determined without additional techniques.

## What techniques can be used to evaluate a complex indeterminate algebraic limit?

Techniques such as L'Hopital's rule, Taylor series, and Cauchy's integral formula can be used to evaluate a complex indeterminate algebraic limit.

## What are some common types of expressions that result in complex indeterminate algebraic limits?

Expressions involving trigonometric functions, exponential functions, and logarithmic functions often result in complex indeterminate algebraic limits.

## Why are complex indeterminate algebraic limits important in mathematics?

Complex indeterminate algebraic limits are important because they arise in many real-world applications, and their evaluation requires advanced mathematical techniques. They also provide a deeper understanding of the behavior of complex numbers and their relationship with algebraic expressions.