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