What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?

Note that\begin{align}\sqrt{n^2+n}-\sqrt[3]{n^3+1}&=\sqrt[6]{(n^2+n)^3}-\sqrt[6]{(n^3+1)^2}\\&=\frac{(n^2+n)^3-(n^3+1)^2}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}\\&=\frac{3n^5+3n^4-n^3-1}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}.\end{align}This quotient behaves as $\dfrac{3n^5}{6n^5}$ and therefore its limit is $\dfrac12$.
