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?

Yes, you can make that approach work. It's a little tedious and messy looking, but here's what you get:
$${\sqrt{n^2+1}-\sqrt{n^2+n}\over\sqrt[3]{n^3+1}-\sqrt[3]{n^3+n^2+1}}\\={(n^2+1)-(n^2+n)\over(n^3+1)-(n^3+n^2+1)}\cdot{(\sqrt[3]{n^3+1})^2+\sqrt[3]{n^3+1}\sqrt[3]{n^3+n^2+1}+(\sqrt[3]{n^3+n^2+1})^2\over\sqrt{n^2+1}+\sqrt{n^2+n}}\\
={n-1\over n^2}\cdot{n^2\left(\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\right)\over n\left(\sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}} \right)}\\
={n-1\over n}\cdot{\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\over \sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}}}\\
\to1\cdot{1^2+1\cdot1+1^2\over1+1}={3\over2}$$
The key is to factor an $n$ out of all the square and cube roots after converting the ratio of differences to a ratio of sums, after which the messy stuff that's left inside each of the roots simply tends to $1$.
Remark: I'm prone to making minus sign mistakes, so I always try to doublecheck that the sign of my answer makes sense. In this case it does: the numerator $\sqrt{n^2+1}-\sqrt{n^2+n}$ is negative, since $n^2+n$ is bigger than $n^2+1$, but so is the denominator, since $n^3+n^2+1$ is bigger than $n^3+1$, hence their ratio is positive, so the limit is non-negative, which agrees with the sign of the answer I got.
