0
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $\tfrac23$.

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

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 $$ \frac{\sqrt[3]{1+t}+\sqrt[3]{1-t}}{t}\cdot\frac{c}{c}=\ldots=\frac{1}{t}\frac{2t}{\sqrt[3]{(1+t)^2}+\sqrt[3]{1-t^2}+\sqrt[3]{(1+t)^2}}\to\frac{2}{3}\quad\text{for }t\to 0. $$

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

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: $$ x(1+x^{-1})^{1/3}-x(1-x^{-1})^{1/3} \approx x(1+(1/3)x^{-1})-x(1-(1/3)x^{-1})=2/3 $$

$\endgroup$
  • $\begingroup$ That may be right, but I have no idea what those symbols mean. yet. $\endgroup$ – Gunz Dec 7 '14 at 21:18
  • 1
    $\begingroup$ If you're interested, check out the generalized binomial theorem and big O notation. $\endgroup$ – p.s. Dec 7 '14 at 21:30
0
$\begingroup$

HINT

I would say

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

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

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

$\endgroup$
-1
$\begingroup$

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*}

$\endgroup$
  • 1
    $\begingroup$ The limit is $2/3$. $\endgroup$ – Aaron Maroja Dec 7 '14 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.