# Limit with square and cube root difference $\lim_{n \to \infty} \left(\sqrt{n^2 + n} - \sqrt{n^3 + n^2}\right)$

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.

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

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.

Hint: $$a^6-b^6=(a-b)(a^5+a^4b+...+ab^4+b^5)$$ The 2nd parenthesis will be your $6$.
$\sqrt{n^2 + n}-\sqrt{n^3 + n^2}= n^\frac12 (n+1)^\frac12-n^\frac23(n+1)^\frac13$
$\lim_{n\to\infty}n\{(1+\frac1n)^\frac12-(1+\frac1n)^\frac13\}$ $=\lim_{n\to\infty}n\{(1+\frac1{2n}+\cdots)-(1+\frac1{3n}+\cdots)\}=\frac12-\frac13=\frac16$