<s>[INST] 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?
 [/INST] 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 $t=1/x$.
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