Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$

Question

I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$ I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$ but it didn't help much. Wolfram says the answer is $\frac{3^{1/2}}{3}$.

Any help would be greatly appreciated.

Answer 1

Using the identity $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$, we get $$ \begin{align} &\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}\\ &=\frac{\left(n^3+n^{1/2}\right)-\left(n^3-1\right)}{\left(n^3+n^{1/2}\right)^{2/3}+\left(n^3+n^{1/2}\right)^{1/3}\left(n^3-1\right)^{1/3}+\left(n^3-1\right)^{2/3}}\cdot\sqrt{3n^3+1}\\ &=\frac{n^{1/2}\left(1+n^{-1/2}\right)}{n^2\left(\left(1+n^{-5/2}\right)^{2/3}+\left(1+n^{-5/2}\right)^{1/3}\left(1-n^{-3}\right)^{1/3}+\left(1-n^{-3}\right)^{2/3}\right)}\cdot\sqrt3n^{3/2}\sqrt{1+\frac1{3n^3}}\\ &=\frac{\overbrace{\left(1+n^{-1/2}\right)}^{\to1}}{\underbrace{\left(1+n^{-5/2}\right)^{2/3}}_{\to1}+\underbrace{\left(1+n^{-5/2}\right)^{1/3}\left(1-n^{-3}\right)^{1/3}}_{\to1}+\underbrace{\left(1-n^{-3}\right)^{2/3}}_{\to1}}\cdot\sqrt3\,\,\overbrace{\sqrt{1+\frac1{3n^3}}}^{\to1}\\ &\to\frac1{\sqrt3} \end{align} $$

Answer 2

With the expansions $$ (n^3+\sqrt{n})^{1/3}\sim n+(1/3)n^{-3/2}+... $$ and $$ (n^3-1)^{1/3}\sim n-(1/3)n^{-2}+... $$ and $$ (3n^3+1)^{1/2}\sim \sqrt{3}n^{3/2}+... $$ as $n\to\infty$, the result follows immediately.

Answer 3

Rewrite, $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$

as $n\left(\left(1+\frac{1}{n^{\frac{5}{2}}} \right)^{\frac{1}{3}} -\left(1-\frac{1}{n^3} \right)^{\frac{1}{3}} \right)\times \sqrt{3}n^{\frac{3}{2}}\left(1+\frac{1}{3n^3}\right)^{\frac{1}{2}}$

Now use the Binomial theorem for fractional powers.

Answer 4

Remember the identity: $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$.

$$\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}$$

$$=\frac{\left(\left(n^3+\sqrt{n}\right)-\left(n^3-1\right)\right)\sqrt{3n^3+1}}{\sqrt[3]{\left(n^3+\sqrt{n}\right)^2}+\sqrt[3]{\left(n^3+\sqrt{n}\right)\left(n^3-1\right)}+\sqrt[3]{\left(n^3-1\right)^2}}$$

$$=\frac{\left(1+\frac{1}{\sqrt{n}}\right)\sqrt{3+\frac{1}{n^3}}}{\sqrt[3]{\left(1+\frac{1}{n^2\sqrt{n}}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2\sqrt{n}}\right)\left(1-\frac{1}{n^3}\right)}+\sqrt[3]{\left(1-\frac{1}{n^3}\right)^2}}$$

$$\stackrel{n\to \infty}\to \frac{(1)\sqrt{3}}{1+1+1}=\frac{1}{\sqrt{3}}$$

Answer 5

Sketch: The mean value theorem shows

$$(n^3+\sqrt n)^{1/3} - (n^3-1)^{1/3} =(1/3)c_n^{-2/3}(\sqrt n +1),$$

where $c_n\in (n^3-1,n^3+\sqrt n).$ Using the endpoints of that interval gives upper and lower bounds on $c_n^{-2/3}.$ Now multipy by $\sqrt {3n^3 + 1},$ pull out the various powers of $n,$ use the squeeze theorem, and the limit becomes apparent.