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?
 A: $$\lim_{n \to \infty} (\sqrt[3]{n^3+1}-\sqrt{n^2+1})=$$
$$=\lim_{n\to \infty}((\sqrt[3]{n^3+1}-n)-(\sqrt{n^2+1}-n))=$$
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.
$$=\lim_{n\to \infty}\left(\frac{(\sqrt[3]{n^3+1})^3-n^3}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{(\sqrt{n^2+1})^2-n^2}{\sqrt{n^2+1}+n}\right)=$$
$$=\lim_{n\to \infty}\left(\frac{1}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{1}{\sqrt{n^2+1}+n}\right)=0-0=0$$
A: 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.
A: Consider
$$
f(x)=\sqrt[3]{1+x^3}-\sqrt{1+x^2}
$$
Then
$$
f'(x)=\frac{3x^2}{3\sqrt[3]{(1+x^3)^2)}}-\frac{2x}{2\sqrt{1+x^2}}
$$
and therefore $f'(0)=0$. Therefore
$$
0=f'(0)=\lim_{x\to0^+}\frac{\sqrt[3]{1+x^3}-\sqrt{1+x^2}}{x}=
\lim_{t\to\infty}\bigl(\sqrt[3]{t^3+1}-\sqrt{t^2+1}\bigr)
$$
with the substitution $t=1/x$.
