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

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?

• Pull out $n^{3/2}$ and use the fact that $(1+x)^{2/3}\sim 1+\frac 2 3x+o(x)$. Commented Feb 4, 2023 at 4:45
• If the expression under limit is $A-B$ then $A/B\to 1$ and $A-B=\frac{(A/B) - 1}{(A/B)^3-1}\cdot\frac{A^3-B^3}{B^2}$. First fraction tends to $1/3$ and the second fraction is simpler to deal as $A^3-B^3=4n$. Commented Feb 4, 2023 at 13:50

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*}
• Indeed. After simplifying the numerator and pulling out n^$\frac{4}{3}$ from the denominator, we see that it's degree is greater than the numerator and we get zero. That's the answer. Commented Feb 4, 2023 at 5:05