<s>[INST] How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$? How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$ ? I know that if it is a square root I will multiply by the conjugate. but if it is a cubic root what shall I do?
Thanks 
 [/INST] Formally substitute $n=1/x^{3/2}$ to get
$$
f(x)=\sqrt[3]{\frac{1}{x^3}+1}-\sqrt[3]{\frac{1}{x^3}+\frac{1}{x^{3/2}}}=
\frac{\sqrt[3]{1+x^3}-\sqrt[3]{1+x^{3/2}}}{x}=
\frac{1-1-x^{3/2}/3+o(x^{3/2})}{x}=-\frac{x^{1/2}}{3}+o(x^{1/2})
$$
Then $\lim_{x\to0^+}f(x)=0$. Since
$$
\sqrt[3]{n^2+1}-\sqrt[3]{n^2+n}=f\left(\frac{1}{n^{2/3}}\right)
$$
you have
$$
\lim_{n\to\infty}f\left(\frac{1}{n^{2/3}}\right)=0
$$
</s>