7. [5] Find $p$ so that $\lim _{x \rightarrow \infty} x^{p}(\sqrt[3]{x+1}+\sqrt[3]{x-1}-2 \sqrt[3]{x})$ is some non-zero real number.
Answer: $\sqrt{\frac{5}{3}}$ Make the substitution $t=\frac{1}{x}$. Then the limit equals to
$$
\lim _{t \rightarrow 0} t^{-p}\left(\sqrt[3]{\frac{1}{t}+1}+\sqrt[3]{\frac{1}{t}-1}-2 \sqrt[3]{\frac{1}{t}}\right)=\lim _{t \rightarrow 0} t^{-p-\frac{1}{3}}(\sqrt[3]{1+t}+\sqrt[3]{1-t}-2) .
$$

We need the degree of the first nonzero term in the MacLaurin expansion of $\sqrt[3]{1+t}+\sqrt[3]{1-t}-2$. We have
$$
\sqrt[3]{1+t}=1+\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right), \quad \sqrt[3]{1-t}=1-\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right) .
$$

It follows that $\sqrt[3]{1+t}+\sqrt[3]{1-t}-2=-\frac{2}{9} t^{2}+o\left(t^{2}\right)$. By consider the degree of the leading term, it follows that $-p-\frac{1}{3}=-2$. So $p=\frac{5}{3}$.