<s>[INST] 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?
 [/INST] Considering $$y= \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$ compose Taylor series and use binomial expansions
$$\sqrt{n^2+1}=n+\frac{1}{2 n}+\cdots$$
$$\sqrt{n^2+n}=n+\frac{1}{2}-\frac{1}{8 n}+\cdots$$
$$\sqrt[3]{n^3+1}=n+\frac{1}{3 n^2}+\cdots$$
$$\sqrt[3]{n^3+n^2+1}=n+\frac{1}{3}-\frac{1}{9 n}+\cdots$$
$$y=\frac {-\frac{1}{2}+\frac{5}{8 n}+\cdots } {-\frac{1}{3}+\frac{1}{9 n}+\cdots  }\to \frac 32$$
</s>