Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multiplicate a polynomial by square and third roots?

Instead, write $\sqrt{n^2+1}= |n|\sqrt{1+\frac{1}{n^2}}$, and similar for other expression in the numerator. For the denominator, multiply both numerator and denominator. Then in the denominator you will have $n^2$, and in the numerator the term
$$
(n^3+1)^{\frac{2}{3}} + ((n^3+1)(n^3+n^2+1))^{\frac{2}{3}} + (n^3+n^2+1)^{\frac{2}{3}}
$$
Then, you can perform the same operation on them, e.g. on the first one it is $|n|(1+\frac{1}{n^3})^{\frac{1}{3}}$, and a lot of terms will cancel out
