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