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I've been banging my head at this one for the last hour and I can't seem to find the solution. I've read through almost all of the limit problems that were asked to be solved here and the tricks used just don't work on this one, or maybe I'm just not seeing it.

$\lim_{n \to \infty} \left(\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}\right)$

Of course the solution has to be done without using the L'Hospital's rule or series expansion.

I've used the $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ identity and $a^2 - b^2 = (a + b)(a - b)$ and got nowhere near the solution -> $\frac{1}{6}$.

I even tried using the $(1 + \frac{1}{n})^n$ and played with exponents but got a solution -> $\dfrac{1 - e^{2/3}}{2}$ which told me I'm definitely doing something wrong.

Martin Sleziak
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    In this case, it might be easier to calculate $\lim_{n\to\infty} (\sqrt{n^2-n}-n)$ and $\lim_{n\to\infty} (\sqrt[3]{n^3-n^2}-n)$ separately. You can find some posts about similar limits here: http://math.stackexchange.com/questions/136495/calculate-lim-n-to-infty-sqrtn2n-n, http://math.stackexchange.com/questions/236901/showing-lim-n-rightarrow-infty-sqrt3n3n2-sqrt3n31-rightarrow-f And if you look at this more general question, you'll find out that several other similar problems are linked there. – Martin Sleziak Oct 09 '16 at 06:20
  • https://math.stackexchange.com/questions/1015889/the-limit-of-a-difference-between-a-cube-root-and-square-root – user236182 Nov 02 '17 at 20:01
  • @MartinSleziak You mean $\lim_{n\to \infty}(\sqrt{n^2+n}-n)$ and $\lim_{n\to \infty}(\sqrt[3]{n^3+n^2}-n)$. – user236182 Nov 02 '17 at 20:10
  • André Nicolas's answer in my linked question gave some new solutions for this question. – user236182 Nov 02 '17 at 20:34

2 Answers2

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Hint: $$a^6-b^6=(a-b)(a^5+a^4b+...+ab^4+b^5)$$ The 2nd parenthesis will be your $6$.

Nameless
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$\sqrt{n^2 + n}-\sqrt[3]{n^3 + n^2}= n^\frac12 (n+1)^\frac12-n^\frac23(n+1)^\frac13$

$\lim_{n\to\infty}n\{(1+\frac1n)^\frac12-(1+\frac1n)^\frac13\}$ $=\lim_{n\to\infty}n\{(1+\frac1{2n}+\cdots)-(1+\frac1{3n}+\cdots)\}=\frac12-\frac13=\frac16$

lab bhattacharjee
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