Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$ Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right) \leq n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
Is there anyone who can give me a hint as to  how to solve it?

$$\lim_{n \to \infty} (\sqrt[3]{n^3+1}-\sqrt{n^2+1})=$$
$$=\lim_{n\to \infty}((\sqrt[3]{n^3+1}-n)-(\sqrt{n^2+1}-n))=$$
Use the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1}), k\ge 2, k\in\mathbb Z, a,b\in\mathbb R$$
with, e.g., $a=\sqrt[3]{n^3+1}$, $b=n$, $k=3$, etc.
$$=\lim_{n\to \infty}\left(\frac{(\sqrt[3]{n^3+1})^3-n^3}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{(\sqrt{n^2+1})^2-n^2}{\sqrt{n^2+1}+n}\right)=$$
$$=\lim_{n\to \infty}\left(\frac{1}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{1}{\sqrt{n^2+1}+n}\right)=0-0=0$$
