Find the limit $\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$

Question

I have tried simplifying in every way I can think of, but I just can't seem to figure this out. I am asked to find the following limit: $$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$$

I don't understand, where does the difference of cubes formula fit here? these are cube roots.. — Rosie, Nov 11 '20 at 16:22
The difference of cubes formula is for $\dfrac{a^3-b^3}{a-b}$. Take $a=\sqrt[3]{n^3+n^2}$ and $b = \sqrt[3]{n^3+1}$. — Robert Israel, Nov 11 '20 at 16:24
See also: Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$. Other questions linked there might be of interest, too. — Martin Sleziak, Feb 27 '21 at 16:25

score 2 · Answer 1 · answered Nov 11 '20 at 16:29

Elementary rearranging gives:

\begin{eqnarray*}&\,\,&\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\\&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n}\\&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n-\frac1{n^3}}\left(1-\frac1{n^2}\right)\\ &=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n-\frac1{n^3}}\lim_{n\to \infty}\left(1-\frac1{n^2}\right)\\ &=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{(1+\frac1n)-(1+\frac1{n^3})} \end{eqnarray*}

Let $f(t)=\sqrt[3](t)$, and let $x_n=1+\frac1n, x'_n=1+\frac1{n^3}$.

Our limit is then:$$ \lim_{n\to \infty} \frac{f(x_n)-f(x'_n)}{x_n-x'_n}=f'(1).$$

Fianlly note $f'(1)=\frac13$.

score 2 · Answer 2 · answered Nov 11 '20 at 16:32

2

You could use binomial series:

$\lim\limits_{n\to \infty}\left[ \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\right]$

$=\lim\limits_{n\to\infty}\left[n\left(1+\dfrac1n\right)^{1/3}-n\left(1+\dfrac1{n^3}\right)^{1/3}\right]$

$=\lim\limits_{n\to\infty}\left[n\left(1+\dfrac{\color{blue}1}{\color{blue}3n}+\cdots\right)-n\left(1+\dfrac1{3n^3}+\cdots\right)\right]=\color{blue}{\dfrac13}$

answered Nov 11 '20 at 16:32

J. W. Tanner

60,406

The answer would be the same if $n^3+1$ were replaced with $n^3+n$ – J. W. Tanner Nov 11 '20 at 16:36

score 2 · Answer 3 · answered Nov 11 '20 at 16:56

As hinted in the comments above: $$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}=\\ \lim_{n\to \infty} \frac{\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}}{1}\cdot \frac{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}=\\ \lim_{n\to \infty} \frac{(n^3+n^2)-(n^3+1)}{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}=\\ \lim_{n\to\infty} \frac{1}{\sqrt[3]{\frac{(n^3+n^2)^2}{(n^2-1)^3}}+\sqrt[3]{\frac{(n^3+n)(n^3+1)}{(n^2-1)^3}}+\sqrt[3]{\frac{(n^3+1)^2}{(n^2-1)^3}}}=\frac1{1+1+1}=\frac13.$$

As the last step, you could divide by $n^2$ instead of $n^2-1$ with a bit less effort & the same result. — lisyarus, Nov 11 '20 at 18:50

Find the limit $\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$

3 Answers3