Finding the limit of $(\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$

Question

Any Ideas/Hints?

$$\lim\limits_{n \to \infty } (\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$$

Answer 1

$$\lim_{n\to\infty}{\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot 1$$ $$=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot \frac{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}$$ $$=\lim_{n\to\infty}\frac{{\sqrt[3]{n^3+n^2}^3}-\sqrt[3]{n^3+1}^3}{{{\sqrt[3]{(n^3+n^2)^2}}++\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}}=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}}$$ $$=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{n^6+2n^5+n^4}}+\sqrt[3]{n^6+n^5+n^3+n^2}+\sqrt[3]{n^6+2n^3+1}}}$$ $$=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{n^6(1+2\frac{1}{n}+\frac{1}{n^2})}}+\sqrt[3]{n^6(1+\frac{1}{n}+\frac{1}{n^3}+\frac{1}{n^4})}+\sqrt[3]{n^6(1+2\frac{1}{n^3}+\frac{1}{n^6})}}}$$ $$=\lim_{n\to\infty}\frac{n^2-1}{\sqrt[3]n^6+\sqrt[3]n^6+\sqrt[3]n^6}=\lim_{n\to\infty}\frac{n^2-1}{n^2+n^2+n^2}=\frac{1}{3}$$

Answer 2

Hint: $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$ and $b=\sqrt[3]{n^3+1}$.

Answer 3

You can extract n from each cube root and use the fact that (1+x)^a is close to (1+a x) when x is small. Then, expand.