Ytukyg

I am having great problems in solving this:

$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$

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:

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

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

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

therefore

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

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

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

Can you take it from here?

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

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?

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