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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?

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  • $\begingroup$ Pull out $n^{3/2}$ and use the fact that $(1+x)^{2/3}\sim 1+\frac 2 3x+o(x)$. $\endgroup$
    – geetha290krm
    Commented Feb 4, 2023 at 4:45
  • $\begingroup$ 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 $A^3-B^3=4n$. $\endgroup$
    – Paramanand Singh ♦
    Commented Feb 4, 2023 at 13:50

1 Answer 1

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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?

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  • $\begingroup$ 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. $\endgroup$
    – StrawberryVanilla
    Commented Feb 4, 2023 at 5:05

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