Prove or disprove $\lim\limits_{n\to\infty} (n+1)^{1/3}−n^{1/3}=\infty$ How can I prove or disprove that $\lim\limits_{n\to \infty} (n+1)^{1/3}−n^{1/3}=\infty$?
My guess is that it is false but I can't prove it.
 A: Let $x^3=n$. Then $$\lim_{n\to\infty}(n+1)^{1/3}-n^{1/3}=\lim_{x\to\infty}(x^3+1)^{1/3}-x.$$
Now for $x\geq1/3$, $$x^3+1\leq x^3+3x\leq x^3+3x+\frac3x+\frac1{x^3}=\left(x+\frac1x\right)^3,$$ so $$\lim_{x\to\infty}(x^3+1)^{1/3}-x\leq\lim_{x\to\infty}\left(\left(x+\frac1x\right)^3\right)^{1/3}-x=\lim_{x\to\infty}\frac1x=0.$$
But $(n+1)^{1/3}-n^{1/3}\geq0$, so $\lim_{n\to\infty}(n+1)^{1/3}-n^{1/3}=0$.
A: The difference of the cube roots of two consecutive positive numbers is always less than the difference of the numbers themselves which is $1$.  So, the limit is not $\infty$.
A: Since $x^3-y^3=(x-y)(x^2+xy+y^2)$, we have
$$
(n+1)^{1/3}-n^{1/3}=\frac{(n+1)-n}{(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}}
$$
Therefore,
$$
\frac1{3(n+1)^{2/3}}\le(n+1)^{1/3}-n^{1/3}\le\frac1{3n^{2/3}}
$$
The limit follows by the Squeeze Theorem.
A: Multiply top and (missing) bottom by $(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}$.
We are exploiting  the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$.  
Another approach is to note that $(n+1)^{1/3}$ is a little less than $n^{1/3}+ \dfrac{1}{3n^{2/3}}$. To prove this, cube the last expression. Then by Squeezing you can see that the limit  of our expression is $0$.
Remark: If all we are interested in is proving that the limit is not "infinity," then it is enough to observe that $(n+1)^{1/3}\lt n^{1/3}+1$ if $n\gt 0$.  To see this, cube both sides. So $(n+1)^{1/3}-n^{1/3}\lt 1$, and therefore cannot get big.
A: Let $f(x)=x^{1/3}$. By mean value theorem, $(n+1)^{1/3}-n^{1/3}=f'(c)=1/(3c^{2/3})\le1/(3n^{2/3})$ for some $c\in(n,\,n+1)$. Hence the difference goes to zero when $n\rightarrow\infty$.
