<s>[INST] 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
 [/INST] $\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}\sqrt[3]{1+1/x} + \sqrt[3]{x}\sqrt[3]{1-1/x} -2\sqrt[3]{x})}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(1+(1/3x)-(1/9x^2) + \cdots) + \sqrt[3]{x}(1-(1/3x)-(1/9x^2) + \cdots) -2\sqrt[3]{x})}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-(1/9x^2) + \cdots) + \sqrt[3]{x}(-(1/9x^2) + \cdots) )}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-2(1/9x^2) + \cdots)) }$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P+(1/3)-2}(((-2/9) + \cdots)) }$
The Higher Negative Powers will tend to $0$ in the limit.
The Only Power term we have is $x^{P+(1/3)-2}$ which must be $x^0$ hence ${P+(1/3)-2} = {0}$
${P+(1/3)-2} = {5/3}$
P less than that will give limit $0$
P more than that will give limit $\infty$
When $P=5/3$ : The Limit is $-2/9$
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