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?

Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$n\left(\left({1+\frac{1}{n^3}}\right)^{1/3}-\left({1+\frac{1}{n^2}}\right)^{1/2}\right)$$
As $n \to \infty$,this can be approximated as
$$n\left(\left({1+\frac{1}{3n^3}+other terms}\right)-\left({1+\frac{1}{2n^2}}+other terms\right)\right)$$
$$n\left({\frac{1}{3n^3}-\frac{1}{2n^2}}+other terms\right)$$
$${\frac{1}{3n^2}-\frac{1}{2n}}+other terms$$
As $n \to \infty$, all terms become 0. So answer is 0.
