# How to solve this limit involving cube root and infinity? $\lim_{x\to\infty} \left((x^3+x^2)^{1/3} - (x^3-x^2)^{1/3})\right)$

How can I solve this limit? I know the answer is $2/3$.

I tried factorisation, but solving the complicated denominator using L'Hopital's Rule returns a wrong answer, $0$.

$$\lim_{x\to\infty} \left((x^3+x^2)^{1/3} - (x^3-x^2)^{1/3})\right)$$

Thanks.

You can use the identity $$a^3−b^3=(a−b)(a^2+ab+b^2)$$ with $a=(x^3+x^2)^{1/3}$ and $b=(x^3-x^2)^{1/3}$. So, you want to multiply (numerator and denominator) with $a^2+ab+b^2$. This will give you \begin{align}a-b&=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}=\frac{a^3-b^3}{a^2+ab+b^2}\\[0.2cm]&=\frac{x^3+x^2-(x^3-x^2)}{(x^3+x^2)^{2/3}+(x^3+x^2)^{1/3}(x^3-x^2)^{1/3}+(x^3-x^2)^{2/3}}\\[0.2cm]&=\frac{2x^2}{(x^{3})^{2/3}(1+\frac1x)+(x^3)^{1/3}(1+\frac1x)(x^3)^{1/3}(1-\frac1x)+(x^3)^{2/3}(1-\frac1x)}\\[0.2cm]&=\frac{2x^2}{x^2(1+\frac1x)+x(1+\frac1x)x(1-\frac1x)+x^2(1-\frac1x)}\\[0.2cm]&=\frac{2}{1+\frac1x+(1+\frac1x)(1-\frac1x)+1-\frac1x}\\[0.2cm]&=\frac{2}{3-\frac1{x^2}}\end{align} Now let $x\to +\infty$ to conclude that $$\lim_{x\to +\infty}\frac{2}{3-\frac1{x^2}}=\frac2{3-0}=\frac23$$

$$\lim_{x\to\infty} (x(1+1/x)^{1/3} - x(1-1/x)^{1/3}))$$ and then expand $(1\pm\epsilon)^{1/3}\approx 1\pm\epsilon/3...$ for $\epsilon\to 0\ .$

With $t:=\frac1x$, The limit is the same as

$$\lim_{t\to0^+}\frac{(1+t)^{1/3}-(1-t)^{1/3}}{t}.$$

By L'Hospital,

$$\lim_{t\to0^+}\frac{(1+t)^{1/3}-(1-t)^{1/3}}{t}=\frac13\lim_{t\to0^+}\left((1+t)^{-2/3}+(1-t)^{-2/3}\right).$$

Alternatively, with the generalized binomial theorem,

$$\lim_{t\to0^+}\frac{\left(1+\frac13t-\frac{1\cdot2}{3\cdot3}\frac{t^2}2+\frac{1\cdot2\cdot5}{3\cdot3\cdot3}\frac{t^3}{3!}\cdots\right)- \left(1-\frac13t-\frac{1\cdot2}{3\cdot3}\frac{t^2}2-\frac{1\cdot2\cdot5}{3\cdot3\cdot3}\frac{t^3}{3!}\cdots\right)}t.$$