# Limit of $\sqrt[3]{n^{3}+n^{2}}-n$.

I tried to proove on my own that $\lim_{n\to\infty} \sqrt[3]{n^{3}+n^{2}}-n=\frac{1}{3}$.

However I couldn't really figure out how to modify the terms so that I could apply the laws for limits.

$$\lim_{n\to\infty}\left( \sqrt[3]{n^{3}+n^{2}}-n\right)=\lim_{n\to\infty}\frac{n^3+n^2-n^3}{\left(\sqrt[3]{n^{3}+n^{2}}\right)^2+n\sqrt[3]{n^{3}+n^{2}}+n^2}=$$ $$=\lim_{n\to\infty}\frac{1}{\left(\sqrt[3]{1+\frac{1}{n}}\right)^2+\sqrt[3]{1+\frac{1}{n}}+1}=\frac{1}{3}$$

You can use the expansion near infinity $$\sqrt[3]{n^3+n^2} =n\sqrt[3]{1+\frac 1n}= n\left(1+\frac 1{3n}+o\left(\frac 1n\right)\right)$$

• How do you get the series at infinity? – Botond Oct 15 '17 at 16:22
• It is just the binomial theorem. For small $x, (1+x)^n \approx 1+nx$ applied to $\frac 1n$, which is small near infinity – Ross Millikan Oct 15 '17 at 16:44
• – user236182 Oct 15 '17 at 20:56

You can shift $n$ by one third and write

$$L=\lim_{n\to\infty} \sqrt[3]{n^{3}-\frac{n}3-\frac1{27}}-n+\frac13.$$

The first two terms will vanish by multiplying/dividing by the conjugate trinomial, because after simplification the numerator is of degree $1$ and the denominator has a term $n^2$

$$n^3+n^2=n^3\left(1+\dfrac1n\right)$$

Set $\sqrt[3]{1+\frac1n}=1+h\implies1+\frac1n=(1+h)^3$

and $h\to0$ as $n\to\infty$

So, we have $$\lim\to0\dfrac h{3h+3h^2+h^3}=?$$

Hint : $$\lim_{x\rightarrow\infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x=\frac{a_{n-1}}{n}$$

Then :

$$\lim_{n}\sqrt[3]{n^{3}+n^{2}}-n=\frac{1}{3}$$

• Funny, I kind of remember having seen the result somewhere... – Did Oct 16 '17 at 12:15