Find the limit of $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ Value of p such that $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ is some finite | non-zero number.
My approach is as follow
$\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {{x^{\frac{{3p}}{3}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$
$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p}}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p + 1}} + {x^{3p}}}} + \sqrt[3]{{{x^{3p + 1}} - {x^{3p}}}} - 2\sqrt[3]{{{x^{3p + 1}}}}} \right)$
How do we proceed

$$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$$
$$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^{p+1/3}}\left( {\sqrt[3]{{1 + 1/x}} + \sqrt[3]{{1 - 1/x}} - 2} \right)} \right)$$
Use $(1+z)^k=1+kz+\frac{k(k-1)}{2}z^2+O(z^3)$ when $z$ is very small, then let $z=1/x$
$$L=\lim_{z\rightarrow 0}~ z^{-p-1/3}\left(1+\frac{z}{3}-\frac{z^2}{9}+O(z^3)+1-\frac{z}{3}-\frac{z^2}{9}+O(z^3)-2\right)$$
$$L=z^{-p-1/3}\left(-\frac{2z^2}{9}+O(z^3)\right)$$
Let $-p-1/3+2=0$, then
$$L=-\frac{2}{9}$$
Hence $p=\frac{5}{3}$ makes the limit finite equal to $-\frac{2}{9}$.
