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?

Generalized binomial theorm:

$(a+b)^k = a^k + ka^{k-1}b + \frac {k(k-1)}{2} a^{k-2}b^2 + \cdots$

$(n^2+n)^\frac 12 = n + \frac 12 - \frac {1}{8} n^{-1} + \cdots\\\ (n^3+1)^\frac 13 = n + \frac 13 n^{-2} + \cdots$

Subtract one from the other and let $n$ go to infinity...

$\frac 12$