<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] First let's get rid of the square root:
$$\lim_{n \to \infty} (\sqrt[3]{n^3-3n^2}-\sqrt{n^2+2n})=\lim_{n \to \infty} \frac{\sqrt[3]{(n^3+3n^2)^2}-(n^2+2n)}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}\\= -\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}.$$
Now let's get rid of the cube root:
$$-\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}$$
$$=-\lim_{n \to \infty} \frac{(n^2+2n)^3-(n^3+3n^2)^2}{(\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n})((n^2+2n)^2+(n^2+2n)\sqrt[3]{(n^3+3n^2)^2}+\sqrt[3]{(n^3+3n^2)^4})}.$$
The numerator yields $-3n^4$ and a bunch of lower-order terms, and the denominator yields $6n^5$ and a bunch of lower order terms, so dividing numerator and denominator by $n^5$ shows that the answer is $0$.
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