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: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$

$\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{
n \left[
\left(
1 + \frac 1n
\right)^{\frac 13} - \left(
1 + \frac 1{n^3}
\right)^{\frac 13}
\right]
\right\} = \\
\displaystyle \lim_{n \rightarrow \infty} \left[
n \left(
1 + \frac 1{3n} - 1 - \frac 1{3n^3}
\right)
\right] = \lim_{n \rightarrow \infty} \left(
\frac 13 - \frac 1{3n^2}
\right) = \frac 13$
