How to evaluate $\lim\limits_{n\rightarrow \infty} [\sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2} ]$

Question

I'm trying to evaluate this limit $\lim\limits_{n\rightarrow \infty} [\sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2} ]$
It is the $\infty-$$\infty$ form. I've tried rewriting it as-

$\lim\limits_{n\rightarrow \infty} [(n+1)^{\frac{2}{3}}-(n-1)^{\frac{2}{3}}]$
and then used the difference of squares formula. Apparantly that only made things worse. I'm stuck with the cube roots. I considered some form of rationalisation but to no avail. How should I approach this limit?

Answer 1

HINT

Multiply and divide by the conjugate in order to obtain: \begin{align*} \sqrt[3]{(n + 1)^{2}} - \sqrt[3]{(n - 1)^{2}} & = \frac{(n + 1)^{2} - (n - 1)^{2}}{\sqrt[3]{(n + 1)^{4}} + \sqrt[3]{(n^{2} - 1)^{2}} + \sqrt[3]{(n - 1)^{4}}} \end{align*}

Can you take it from here?