Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
 A: You should use that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$, $b=\sqrt[3]{n^3+1}$ and then multiply your expression by $(a^2+ab+b^2)/(a^2+ab+b^2)$. Then use the trick you are trying to use.
A: $\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{
n \left[
\left(
1 + \frac 1n
\right)^{\frac 13} - \left(
1 + \frac 1{n^3}
\right)^{\frac 13}
\right]
\right\} = \\
\displaystyle \lim_{n \rightarrow \infty} \left[
n \left(
1 + \frac 1{3n} - 1 - \frac 1{3n^3}
\right)
\right] = \lim_{n \rightarrow \infty} \left(
\frac 13 - \frac 1{3n^2}
\right) = \frac 13$
A: My answer here
(Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $)
shows that
$\sqrt[a]{n^a+n^{a-c}}
=n+\dfrac{1}{an^{c-1}}+O(n^{-(2c-1)})
$.
If $a=3, c=1$,
$\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}+O(n^{-1})
$.
If $a=3, c=3$,
$\sqrt[3]{n^3+1}
=n+\dfrac{1}{3n^2}+O(n^{-5})
=n+O(n^{-2})
$.
Their difference is,
therefore,
$\frac1{3}+O(n^{-1})
\to \frac13
$.
