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.
 A: $$
\begin{align}
\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^{1/3}-\left(1+\frac1{n^3}\right)^{1/3}\right)\\
&=\lim_{n\to\infty}n\left(\left[1+\frac1{3n}+O\!\left(\frac1{n^2}\right)\right]-\left[1+O\!\left(\frac1{n^3}\right)\right]\right)\\
&=\lim_{n\to\infty}\left(\frac13+O\!\left(\frac1n\right)\right)\\
&=\frac13
\end{align}
$$
A: 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 $g'(0)=1/3$.
A: 1) $n(1+1/n )^{1/3} = $
$\dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$\dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$\small{\dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$\small{=\dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)\dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$\lim_{n \rightarrow \infty}\dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$\small{-\lim_{n \rightarrow \infty}(1/n^2)\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$\small{\lim_{n \rightarrow \infty}(1/n^2)×}$
$\small{\lim_{n \rightarrow \infty}\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$\small{0 \cdot (1/3)=0.}$
