# Difficult limit problem

#### sonya36

I have this question:

lim (x^3+x^2)^(1/3)-(x^3-X^2)^(1/3)
x->infinity

I know that I have to simplify this equation first into a fractional form before applying L'Hopital's rule since it is in indeterminate form (infinity - infinity).
I tried to get rid of the cube roots by multiplying by a^2+ab+b^2 but it doesn't seem to work.
I have also tried to separate the limit in to two parts by lim(x^3+x^2)^(1/3) - lim(x^3-X^2)^(1/3) but I am still completely stuck after that.
It seems like it gets more complicated when you try to simplify it.

Can any of you help me solve this problem? Thanks.

#### Jester

MHF Helper
This is what I would try. First factor an x from each term so we have

$$\displaystyle \lim_{x \to \infty} x \left(1 + \dfrac{1}{x}\right)^{1/3} - x \left(1 - \dfrac{1}{x}\right)^{1/3}$$

or

$$\displaystyle \lim_{x \to \infty} x\left[ \left(1 + \dfrac{1}{x}\right)^{1/3} - \left(1 - \dfrac{1}{x}\right)^{1/3}\right]$$

If you let $$\displaystyle a = 1 + \dfrac{1}{x}$$ and $$\displaystyle b = 1 - \dfrac{1}{x}$$ then multiply top and bottom by $$\displaystyle a^2 + ab + b^2$$ and simplify.

See where that takes you.

• 1 person

#### HallsofIvy

MHF Helper
I have this question:

lim (x^3+x^2)^(1/3)-(x^3-X^2)^(1/3)
x->infinity

I know that I have to simplify this equation first into a fractional form before applying L'Hopital's rule since it is in indeterminate form (infinity - infinity).
I tried to get rid of the cube roots by multiplying by a^2+ab+b^2 but it doesn't seem to work.
I presume you mean multiply by the fraction $$\displaystyle \frac{a^2+ ab+ b^2}{a^2+ ab+ b^2}= \frac{(x^3+ x^2)^{2/3}+ (x^3+ x^2)^{1/3}(x^3- x^2)^{1/3}+ (x^3- x^2)^{2/3}}{(x^3+ x^2)^{2/3}+ (x^3+ x^2)^{1/3}(x^3- x^2}^{1/3}+ (x^3- x^2)^{2/3}}$$ so that you get $$\displaystyle (x^3+ x^2)- (x^3- x^2)= 2x^2$$. That sounds good to me! What did you get when you tried that?

Since the limit is as x is going to infinity, you will want to divide both numerator and denominator by the highest power of x so that you will have fractions with x in the denominator. Here that is $$\displaystyle x^2\(\displaystyle so the numerator will be \(\displaystyle x^2$$. In the denominator, you will have to take $$\displaystyle x^2$$ inside each or the parentheses. You will need to use $$\displaystyle x^{-2}(\_\_\_)^{2/3}= (x^{-2(3/2}(\_\_\_))= (x^{-3}(\_\_$$\) and $$\displaystyle x^{-2}\_\_\_)^{1/3}= (x^{-2(3/1)}(\_\_\_))= (x^{-6}(\_\_\_))$$

I have also tried to separate the limit in to two parts by lim(x^3+x^2)^(1/3) - lim(x^3-X^2)^(1/3) but I am still completely stuck after that.
It seems like it gets more complicated when you try to simplify it.

Can any of you help me solve this problem? Thanks.
\)\)

• 1 person

#### JeffM

One way to get this into fractional form is

$(x^3 + x^2)^{1/3} - (x^3 - x^2)^{1/3} = \sqrt{x^3 + x^2} - \sqrt{x^3 - x^2} = y - z = y - z * \dfrac{y^2 + yz + z^2}{y^2 + yz + z^2} = \dfrac{y^3 - z^3}{y^2 + yz + z^2} =$

$\dfrac{\left(\sqrt{x^3 + x^2}\right)^3 - \left(\sqrt{x^3 - x^2}\right)^3}{\left(\sqrt{x^3 + x^2}\right)^2 + \left(\sqrt{x^3 + x^2}\right)\left(\sqrt{x^3 - x^2}\right) + \left(\sqrt{x^3 - x^2}\right)^2} = \dfrac{(x^3 + x^2) - (x^3 - x^2)}{\sqrt{x^6 + 2x^5 + x^4} + \sqrt{x^6 - x^4} + \sqrt{x^6 - 2x^5 + x^4}} =$

$\dfrac{2x^2}{\sqrt{x^3(x^3 + 2x^2 + x)} + \sqrt{x^3(x^3 - x)} + \sqrt{x^3(x^3 - 2x^2 + x)}} = \dfrac{2x^2}{x\left(\sqrt{x^3 + 2x^2 + x} + \sqrt{x^3 - x} + \sqrt{x^3 - 2x^2 + x}\right)} =$

$\dfrac{2x}{\sqrt{x^3 + 2x^2 + x} + \sqrt{x^3 - x} + \sqrt{x^3 - 2x^2 + x}}.$

But I am not sure that that is the form you need to proceed. You can give it a try though (assuming I did not screw up the algebra).

• 1 person

#### sonya36

It worked! I got 2/3, which is the correct answer. Thank you!

#### Soroban

MHF Hall of Honor
Hello, sonya36!

$$\displaystyle \lim_{x\to\infty} (x^3+x^2)^{\frac{1}{3}}-(x^3-x^2)^{\frac{1}{3}}$$

Following up on HallsofIvy's work . . .

Multiply by $$\displaystyle \frac{(x^3+x^2)^{\frac{2}{3}} + (x^3+x^2)^{\frac{1}{3}}(x^3-x^2)^{\frac{1}{3}} + (x^3-x^2)^{\frac{2}{3}}}{(x^3+x^2)^{\frac{2}{3}} + (x^3+x^2)^{\frac{1}{3}}(x^3-x^2)^{\frac{1}{3}} + (x^3-x^2)^{\frac{2}{3}}}$$

and we get: .$$\displaystyle \frac{(x^3+x^2) - (x^3 - x^2)}{(x^3+x^2)^{\frac{2}{3}} + (x^3+x^2)^{\frac{1}{3}}(x^3-x^2)^{\frac{1}{3}} + (x^3-x^2)^{\frac{2}{3}}}$$

. . . . . . . . $$\displaystyle =\;\frac{2x^2}{(x^3+x^2)^{\frac{2}{3}} + (x^6-x^4)^{\frac{1}{3}} + (x^3-x^2)^{\frac{2}{3}}}$$

Divide numerator and denominator by $$\displaystyle x^2$$

. . $$\displaystyle \frac{\dfrac{2x^2}{x^2}} {\dfrac{(x^3+x^2)^{\frac{2}{3}}}{x^2} + \dfrac{(x^6-x^4)^{\frac{1}{3}}}{x^2} + \dfrac{(x^3-x^2)^{\frac{2}{3}}}{x^2}} \;=\;\frac{2}{\left(\dfrac{x^3+x^2}{x^3}\right)^{ \frac{2}{3}} + \left(\dfrac{x^6-x^4}{x^6}\right)^{\frac{1}{3}} + \left(\dfrac{x^3-x^2}{x^3}\right)^{\frac{2}{3}}}$$

Therefore: .$$\displaystyle \lim_{x\to\infty}\frac{2}{\left(1 + \frac{1}{x}\right)^{\frac{2}{3}} + \left(1 - \frac{1}{x^2}\right)^{\frac{1}{3}} + \left(1 - \frac{1}{x}\right)^{\frac{2}{3}}} \;=\;\frac{2}{1+1+1} \;=\;\frac{2}{3}$$

• 1 person