<s>[INST] Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
 [/INST] For positive $n$, let $a=\sqrt[3]{n^3+3n^2}$, and let $b=\sqrt{n^2+2n}$.

Note that $a < n+1$ and
\begin{align*}
3n +1 &= (n+1)^3-a^3\\[4pt]
&= \bigl((n+1)-a\bigr)\bigl((n+1)^2+a(n+1)+a^2)\\[4pt]
&> \bigl((n+1)-a\bigr)(n+1)^2\\[4pt]
\end{align*}
hence $0 < n+1-a < {\large{\frac{3n+1}{(n+1)^2}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-a = 0$.

Also, $b < n+1$ and
\begin{align*}
1&=(n+1)^2-b^2\\[4pt]
&= \bigl((n+1)-b\bigr)\bigl((n+1)+b\bigr)\\[4pt]
&> \bigl((n+1)-b\bigr)(n+1)\\[4pt]
\end{align*}
hence $0 < (n+1)-b < {\large{\frac{1}{n+1}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-b = 0$.

Then we get
$$\lim_{n\to \infty}\;a-b = \lim_{n\to \infty}\;\bigl(a - (n+1)\bigr) + \bigl((n+1)-b\bigr) = 0 + 0 = 0$$ 
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