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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