Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$ Clearly, $\infty -\infty \not =0$
I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?
 A: Let $f:[n-1,n+1]\to\mathbb{R}$ and $f(x)=x^{\frac{2}{3}}$. Then from Mean value theorem (Lagrange’a) exist $\xi\in (n-1,n+1)$ such that:
$$\frac{f(n+1)-f(n-1)}{n+1-(n-1)}=f'(\xi)$$
but (check it out)
$$\frac{2}{3 \sqrt[3]{n+1} }\le f'(\xi)\le \frac{2}{3 \sqrt[3]{n-1} } $$
so
$$\frac{4}{3 \sqrt[3]{n+1} }\le f(n+1)-f(n-1 )\le \frac{4}{3 \sqrt[3]{n-1} } $$
$$\frac{4}{3 \sqrt[3]{n+1} }\le (n+1)^{\frac{2}{3}}-(n-1 )^{\frac{2}{3}}\le \frac{4}{3 \sqrt[3]{n-1} } $$
now You can aplay Squeeze theorem.
A: $n=\frac{1}{t}$ $$\lim_{t\to 0}\frac{{(1+t)}^{2/3}-{(1-t)}^{2/3}}{t^{2/3}}$$  Now apply L-Hospitals rule.. to get $$\lim_{t\to 0}\left(\frac{t}{\sqrt[3]{1+t}}+\frac{t}{\sqrt[3]{1-t}}\right)=?$$
A: $$(n+1)^{2/3} -(n-1)^{2/3}=\frac{(n+1)^2-(n-1)^2}{(n+1)^{4/3}+(n+1)^{2/3}(n-1)^{2/3}+(n-1)^{4/3}}
\\=\frac{n}{n^{4/3}}\frac4{(1+n^{-1})^{4/3}+(1-n^{-2})^{2/3}+(1-n^{-1})^{4/3}}\to 0$$
A: Your expression almost looks like the symetric derivative of $f(x)=x^{2/3} $, so you have :
$$(n+1)^{2/3}-(n-1)^{2/3}\sim_{\infty}2\times\frac{2}{3}n^{-1/3} $$
Therefore, $\lim_{n\to\infty}(n+1)^{2/3}-(n-1)^{2/3}=0 $
edit : $$(n+1)^{2/3}-(n-1)^{2/3}= n^{2/3}\left(\left(1+\frac{1}{n}\right)^{2/3} -\left(1-\frac{1}{n}\right)^{2/3}\right)\\
=n^{2/3}\left(1+\frac{2}{3n}+o\left(\frac{1}{n}\right) -1+\frac{2}{3n}+o\left(\frac{1}{n}\right)\right)\\
=\frac{4}{3}n^{-1/3}+o\left(n^{-1/3}\right) $$
A: Set $\dfrac1n=y\implies y\to0^+$ to find
$$F=\lim_{y\to0^+}\dfrac{(1+y)^{2/3}-(1-y)^{2/3}}y$$
Now rationalize the numerator using $a^3-b^3=(a-b)\cdot (?)$
to find
$$F=\lim_{y\to0^+}\dfrac{(1+y)^2-(1-y)^2}y\cdot\lim_{y\to0^+}\dfrac1{(1+y)^{4/3}+(1+y)^{2/3}(1-y)^{2/3}+(1+y)^{4/3}}=?$$
A: Using Taylor expansion
$$(1+\epsilon)^{2\over 3}= 1+{2\over 3}\epsilon +o(\epsilon)$$
We can write with $\epsilon ={1\over n}$
$$(n+1)^{2\over 3}-(n-1)^{2\over 3}=n^{2\over 3}\left(1+{2\over 3}\epsilon-1+{2\over 3}\epsilon+o(\epsilon)\right)={4\over 3}n^{-{1\over 3}}+o\left(n^{-{1\over 3}}\right)$$
And this quantity at $+\infty$ has limit $0$
