# How to evaluate the following limit? $\lim\limits_{x\to \infty}\big(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\big)$

How to evaluate the following limit? $$\lim _{x\to \infty }\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right).$$

What I first did is multiply by the conjugate, but having trouble finishing the problem. I believe the final answer is $\tfrac23$.

Edit: Here's what I got so far

$$\lim _{x\to \infty }\left(\frac{2x^2}{\sqrt[3]{x^3+x^2}+\sqrt[3]{x^3-x^2}}\right)$$

Recall that $$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right).$$ By taking $a=\sqrt[3]{x^3+x^2}$ and $b=\sqrt[3]{x^3-x^2}$ we get \eqalign{\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right)\left(\sqrt[3]{x^3+x^2}^2+\sqrt[3]{x^3+x^2}\sqrt[3]{x^3-x^2}+\sqrt[3]{x^3-x^2}^2\right)=\sqrt[3]{x^3+x^2}^3-\sqrt[3]{x^3-x^2}^3. } Hence \eqalign{ \sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}=\dfrac{2x^2}{\sqrt[3]{x^3+x^2}^2+\sqrt[3]{x^3+x^2}\sqrt[3]{x^3-x^2}+\sqrt[3]{x^3-x^2}^2} .} Now divide top and bottom by $\tfrac1{x^2}$ to find \eqalign{\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}= \dfrac{2}{\sqrt[3]{\tfrac1x+1}^{2}+\sqrt[3]{\tfrac{-1}{x}+1}+\sqrt[3]{\tfrac{-1}x+1}^{2}}, } thus the limit as $x\to\infty$ is as expected equal to $\tfrac23$.

• I stated in the description of multiplying by the conjugate. After that is where I'm having trouble finishing the problem – Gunz Dec 7 '14 at 19:09
• @Gunz Look at my edit. – Hakim Dec 7 '14 at 20:06
• wouldn't x^3 / x^2 = x tho, not 1/x? – Gunz Dec 7 '14 at 21:19
• @Gunz No, one has \eqalign{\sqrt[3]{x^3+x^2}^2\cdot\dfrac1{x^2}&=\dfrac{\sqrt[3]{x^3+x^2}^2}{x^2}\\&=\left(\dfrac{\sqrt[3]{x^3+x^2}}{x}\right)^2\\&=\left( \dfrac {\sqrt[3]{x^3+x^2}}{\sqrt[3]{x^3}}\right)^2\\&=\left(\sqrt[3]{\dfrac{x^3+x^2}{x^3}}\right)^2\\&=\left(\sqrt[3]{\dfrac1x+1}\right)^2\\&=\sqrt[3]{\dfrac1x+1}^2.} – Hakim Dec 7 '14 at 21:56
• Ah i see. Thank you very much. – Gunz Dec 7 '14 at 22:13

Put $t=\frac{1}{x}$, so that the limit $L=\lim _{x\to \infty }\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right)$ becomes $$L=\lim_{t\to 0}\frac{\sqrt[3]{1+t}-\sqrt[3]{1-t}}{t}.$$ observing that $a^{1/3}-b^{1/3}=\left(a-b\right)\underbrace{\left(a^{2/3}+\sqrt[3]{ab}+b^{2/3}\right)}_c$ with $a=1+t$ and $b=1-t$ multiplying and dividing by $c$, we have $$\frac{\sqrt[3]{1+t}+\sqrt[3]{1-t}}{t}\cdot\frac{c}{c}=\ldots=\frac{1}{t}\frac{2t}{\sqrt[3]{(1+t)^2}+\sqrt[3]{1-t^2}+\sqrt[3]{(1+t)^2}}\to\frac{2}{3}\quad\text{for }t\to 0.$$

• I think the $+$ should be a $-$ in both instances of the expression $\frac{\sqrt[3]{1+t}+\sqrt[3]{1-t}}{t}$. – Théophile Dec 7 '14 at 20:42

Any time you want to evaluate a difference of functions which are asymptotic to each other, I find the most straightforward approach is to expand each as an asymptotic series and then look at the noncanceling terms. In this case:

\begin{aligned} (x^3+x^2)^{1/3}-(x^3-x^2)^{1/3}&=x(1+x^{-1})^{1/3}-x(1-x^{-1})^{1/3}\\ &=x \sum_{k=0}^\infty\binom{1/3}{k}x^{-k}-x \sum_{k=0}^\infty\binom{1/3}{k}(-x)^{-k}\\ &=(x + 1/3 + O(x^{-1})) - (x-1/3+O(x^{-1}))\\ &=2/3 + O(x^{-1}) \end{aligned}

Less formally, we just need the approximation $(1+t)^\alpha\approx1+\alpha t$ for $t$ near $0$. So for large $x$, the quantity $x^{-1}$ is near $0$, so we have: $$x(1+x^{-1})^{1/3}-x(1-x^{-1})^{1/3} \approx x(1+(1/3)x^{-1})-x(1-(1/3)x^{-1})=2/3$$

• That may be right, but I have no idea what those symbols mean. yet. – Gunz Dec 7 '14 at 21:18
• If you're interested, check out the generalized binomial theorem and big O notation. – p.s. Dec 7 '14 at 21:30

HINT

I would say

$a=\sqrt[3]{x^3+x^2},\quad b=\sqrt[3]{x^3-x^2}$

$\displaystyle (a-b)\cdot \frac{a^2+ab+b^2}{a^2+ab+b^2}=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{2x^2}{a^2+ab+b^2}$

$\displaystyle \lim _{x\to\infty}\frac{2x^2}{a^2+ab+b^2}=\lim _{x\to\infty}\frac{2x^2}{(x^3+x^2)^{2/3}+(x^3+x^2)^{1/3}(x^3-x^2)^{1/3}+b^2+(x^3-x^2)^{1/3}}=\cdots$

If you use $(\cdots)^{1/3}+(\cdots)^{1/3}$ as the "conjugate" you cannot get $2x^2$ on the numerator. \begin{align*} \lim_{x\rightarrow\infty}(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3+x^2})&=\lim_{x\rightarrow\infty}\frac{(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2})((x^3+x^2)^{2/3}+(x^3-x^2)^{2/3})}{(x^3+x^2)^{2/3}+(x^3-x^2)^{2/3}}\\ &=\lim_{x\rightarrow\infty}\frac{x^3+x^2-x^3+x^2}{(x^3+x^2)^{2/3}+(x^3-x^2)^{2/3}}\\ &=\lim_{x\rightarrow\infty}\frac{2x^2}{x^2(1+x^2/x^3)^{2/3}+x^2(1-x^2/x^3)^{2/3}}\\ &=\lim_{x\rightarrow\infty}\frac{2}{(1+1/x)^{2/3}+(1-1/x)^{2/3}}\\ &=1. \end{align*}

• The limit is $2/3$. – Aaron Maroja Dec 7 '14 at 19:39