<s>[INST] Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt[3]{(n^3+2n^2+1)^2} + \sqrt[3]{(n^3+2n^2+1)(n^3-1)} + \sqrt[3]{(n^3-1)^2}}$$
$$\lim\limits_{n \to \infty} \dfrac{2n^2+2}{\sqrt[3]{n^6+4n^5+4n^4+2n^3+4n^2+1} + \sqrt[3]{n^6+2n^5-2n^2-1} + \sqrt[3]{n^6-2n^3+1}}$$
And I saw that we can factor $n^2$ in the denominator and if we do the same in the numerator, we'd get that the limit is equal to $2$. The problem is that my textbook claims that this limit is actually $\dfrac{2}{3}$. I don't see why should I have a $3$ in the denominator since the coefficient of $n^2$ would be $1$ if I would carry out the factoring to detail. So, what did I do wrong?
 [/INST] Since
$(n+1)^3
=n^3+3n^2+$lower order terms,
$(n+2/3)^3
=n^3+2n^2+$lower order terms.
Therefore the left cube root
is about $n+2/3$.
Since the right cube root
is just about $n$,
the difference is about $2/3$.
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