Solve $\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$

Question

I am having great problems in solving this:

$$\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$$

I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so

$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$

I tried expanding it as well, which led to absolutely nothing. These are my writings to this:

Answer 1

$\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{n+\sqrt{n}-n}{(n+\sqrt{n})^{2/3}+(n+\sqrt{n})^{1/3}n^{1/3}+n^{2/3}}\leq \frac{\sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=\frac{\sqrt{n}}{3n^{2/3}}\rightarrow 0$

Answer 2

Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$

We have that $$0< \sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{\sqrt[3]{n}}{\sqrt{n}}\cdot\frac{ \sqrt[3]{1+\frac{1}{\sqrt{n}}}-1}{\frac{1}{\sqrt{n}}}\leq \frac{1}{\sqrt[6]{n}}\cdot \frac{1}{3}.$$ where we used the Bernoulli's inequality $$(1+x)^r\leq 1+rx$$ with $r=1/3\in(0,1)$ and $x=1/\sqrt{n}$.

Can you take it from here?

P.S. By using this approach we are also able to find
$$\lim\limits_{n\to\infty}\left((n+\sqrt{n})^r-n^r\right)$$ for any $r<1/2$. Note that if we replace $\sqrt[3]{\ }$ with say $\sqrt[5]{\ }$ then the "algebraic way" could be very annoying!

Answer 3

As you suggested, \begin{align} \sqrt[3]{n+\sqrt n} - \sqrt[3]{n} &= \frac{n+\sqrt n - n}{(n+\sqrt n)^{2/3}+\sqrt[3]{n(n+\sqrt n)} + n^{2/3}}\\ &= \frac{\sqrt n}{(n+\sqrt n)^{2/3}+\sqrt[3]{n(n+\sqrt n)} + n^{2/3}}. \end{align} The fastest growing term is evidently $n^{2/3}$. Can you finish?

Answer 4

By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have

$$\sqrt[3]{n+\sqrt{n}}=\sqrt[3]{n}\, \left(1+\frac1{\sqrt n}\right)^\frac13=\sqrt[3]{n}+\frac{\sqrt[3]{n}}{3\sqrt n}+o\left(\frac{\sqrt[3]{n}}{\sqrt n}\right)=\sqrt[3]{n}+\frac{1}{3n^\frac16}+o\left(\frac{1}{n^\frac16}\right)$$

therefore

$$\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{1}{3n^\frac16}+o\left(\frac{1}{n^\frac16}\right)$$

Answer 5

Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $y\in (n, n+\sqrt n)$ such that $$ f(n+\sqrt n) - f(n) = f'(y)(n+\sqrt n - n)= \frac{y^{-2/3}}{3}\sqrt n<n^{-2/3}\sqrt n=n^{-1/6}\to 0. $$