Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$\frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}\longrightarrow\frac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.

My favorite way is to consider
$$
f(x)=\left(\frac{1}{x^3}+\frac{1}{x^2}\right)^{1/3}-\left(\frac{1}{x^3}+1\right)^{1/3}
=\frac{\sqrt[3]{1+x}-\sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
\lim_{x\to0^+}f(x)=\lim_{x\to0}\frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=\frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=\sqrt[3]{1+x}-\sqrt[3]{1+x^3}$, so
$$
g'(x)=\frac{1}{3\sqrt[3]{(1+x)^2}}-\frac{3x^2}{3\sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
