Expanding Cube Roots: Solving Limits with Maclaurin Series

In summary, the conversation discusses a problem from Steven Krantz's Calculus Demystified where the goal is to find the limit of a complex function. One person suggests using the Maclaurin series to solve it, while another suggests using high school algebra. They also discuss a solution using a simple substitution.
  • #1
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A quiz at the end of Steven Krantz's Calculus Demystified includes the following problem:
Find
[tex]
\lim_{x \to \infty} [ \sqrt[3]{x+1}
-
\sqrt[3]{x} ]
[/tex]
I see how one can use the Maclaurin series to get
[tex]
\sqrt[3]{x+1} = 1 + \frac {x}{3} - \frac {x^2}{9} + \frac {5 x^3}{81} + . . .
[/tex]
but trying it with the cube root of x gives me zero plus an endless series of undefined terms.
Is there a way to expand [itex] \sqrt[3]{x} [/itex] and solve this problem?
 
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  • #2
But why do you want expand that in series?
Isn't it by simple high school algebra :
ac{2}{3}+%20%28x+1%29^\frac{2}{3}+x^{\frac{1}{3}}%28x+1%29^{\frac{1}{3}}%20}.gif

for x>0?
 
  • #3
Oh yeah, you're right zoki85 - thanks!

I also just realized that I could simply define y = x + 1, plug in y-1 for all the x's in the series above and there's the answer to my other question. :)
 

Related to Expanding Cube Roots: Solving Limits with Maclaurin Series

What is a cube root?

A cube root is a mathematical operation that finds the number which, when multiplied by itself three times, gives the given number as a result. It is the inverse of cubing a number.

How do you expand a simple cube root?

To expand a simple cube root, you can use the property of cube roots which states that the cube root of a product is equal to the product of the cube roots of each individual factor. This can be written as:
$$\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$$

What are some common mistakes when expanding a simple cube root?

One common mistake is to try to expand the cube root of a sum, which cannot be done using the property mentioned above. Another mistake is to forget to simplify the cube root of perfect cube numbers.

Can you expand a cube root with variables?

Yes, the same property of cube roots applies to variables as well. However, it is important to note that the variables must have an exponent that is a multiple of three in order to be able to simplify the cube root fully.

Why is expanding a simple cube root useful?

Expanding a simple cube root can be useful in solving equations and simplifying expressions involving cube roots. It can also help in understanding the relationships between numbers and their cube roots.

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