Question

If $\underset{x\to \infty }{lim}\left(\right(x^3+x^2{)}^{1/3}-(x^3-x^2{)}^{1/3})=R$ then $3R=$

Answer 1

$\underset{x\to \infty }{lim}\left(\right(x^3+x^2{)}^{1/3}-(x^3-x^2{)}^{1/3})$
We know that $a^3-b^3=(a-b)(a^2+ab+b^2)$
Let $a=(x^3+x^2{)}^{1/3}$ and $b=(x^3-x^2{)}^{1/3}$

$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$
$\Rightarrow \underset{x\to \infty }{lim}\left(\right(x^3+x^2{)}^{1/3}-(x^3-x^2{)}^{1/3})$
$=\underset{x\to \infty }{lim}\frac{x^3+x^2-x^3+x^2}{(x^3+x^2{)}^{2/3}+(x^3+x^2{)}^{1/3}(x^3-x^2{)}^{1/3}+(x^3-x^2{)}^{2/3}}$

$=\underset{x\to \infty }{lim}\frac{2x^2}{x^2{(1+\frac{1}{x})}^{2/3}+x{(1+\frac{1}{x})}^{1/3}x{(1-\frac{1}{x})}^{1/3}+x^2{(1-\frac{1}{x})}^{2/3}}$

$=\underset{x\to \infty }{lim}\frac{2}{{(1+\frac{1}{x})}^{2/3}+{(1+\frac{1}{x})}^{1/3}{(1-\frac{1}{x})}^{1/3}+{(1-\frac{1}{x})}^{2/3}}$

$=\frac{2}{{(1+0)}^{2/3}+{(1+0)}^{1/3}{(1-0)}^{1/3}+{(1-0)}^{2/3}}$
$=\frac{2}{3}=R$

$3R=2$