Difficult limit problem

Nov 2014
26
0
Canada
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
Dec 2008
2,470
1,255
Conway AR
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.
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
\)\)
 
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Feb 2014
1,748
651
United States
One way to get this into fractional form is

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

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

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

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

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).
 
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Nov 2014
26
0
Canada
It worked! I got 2/3, which is the correct answer. Thank you!
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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: .\(\displaystyle \lim_{x\to\infty}\frac{2}{\left(1 + \frac{1}{x}\right)^{\frac{2}{3}} + \left(1 - \frac{1}{x^2}\right)^{\frac{1}{3}} + \left(1 - \frac{1}{x}\right)^{\frac{2}{3}}} \;=\;\frac{2}{1+1+1} \;=\;\frac{2}{3}\)

 
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