In need of assistance with evaluation of a tricky limit I am trying to evaluate a limit, but without much luck.
I keep getting infinity as the answer, but both Wolfram and my textbook state otherwise.
The limit problem goes as follows:
$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}}) $ 
/* EDIT */
This is my original work, I wonder why it has failed to yield the result of zero as expected.

Please solve it step by step, in order for me to understand my error.
Thank you!
 A: You'll need to use Taylor's expansion at order $2$. Rewrite your expression as 
\begin{align*}& n^{\tfrac23+\tfrac12}\Bigl(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}-2\Bigr)\\
{}={}& n^{\tfrac76}\biggl(1+\frac1{2n}-\frac1{8n^2}+o\Bigl(\frac1{n^2}\Bigr)+1-\frac1{2n}-\frac1{8n^2}+o\Bigl(\frac1{n^2}\Bigr)-2\biggl)\\
{}={}& n^{\tfrac76}\biggl(-\frac1{4n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggl)=-\frac1{4n^{\tfrac56}}+o\Biggl(\frac1{n^{\tfrac56}}\Biggr)
\end{align*}
Hence the limit is $0$.
A: A difference of radicals can be simplified by multiplying "top and bottom" by its conjugate:
$$\sqrt a-\sqrt b = (\sqrt a - \sqrt b){\sqrt a +\sqrt b\over\sqrt a+\sqrt b}={a-b\over\sqrt a+\sqrt b}.$$
Observing that your parenthesized expression $P:=(\sqrt{n+1}+\sqrt{n-1}-2\sqrt n)$ can be written $P=(\sqrt{n+1}-\sqrt n)+(\sqrt{n-1}-\sqrt n)$,
use the conjugate device twice to get
$$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\;\;\mbox{and}\;\;\sqrt{n-1}-\sqrt n=\frac{-1}{\sqrt{n-1}+\sqrt n}.$$
Adding, we get
$$P={(\sqrt{n-1}+\sqrt n)-(\sqrt{n+1}+\sqrt n)\over(\sqrt{n-1}+\sqrt n)(\sqrt{n+1}+\sqrt n)}=
{(\sqrt{n-1}-\sqrt{n+1})\over(\sqrt{n-1}+\sqrt n)(\sqrt{n+1}+\sqrt n)}$$
and applying the conjugate device one more time gives (after simplification)
$$
P={-2\over(\sqrt{n-1}+\sqrt n)(\sqrt{n+1}+\sqrt n)(\sqrt{n-1}+\sqrt{n+1})}.
$$
Conclude that $P$ behaves like $n^{-3/2}$ as $n\to\infty$, so $n^{2/3}P$ behaves like $n^{-5/6}$, and the limit is zero.
A: $$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})=$$
$$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})\frac{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{({\sqrt{n+1}}+{\sqrt{n-1}})^2-4n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{2\sqrt{(n+1)(n-1)}+2n}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{4(n+1)(n-1)-4n^2}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{1}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty} n^{2/3}\frac{-4}{(\sqrt{(n+1)(n-1)}+2n)({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=$$
$$\lim_{n \to \infty} \frac{-4}{(\sqrt{(n^{1/3}+n^{-2/3})(n^{1/3}-n^{-2/3})}+2n^{1/3})({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=0$$
Misread $\frac{2}{3}$ as $\frac{3}{2}$ in the original post 
A: This is of the form $0\cdot\infty$. In these circumstance, when calculating $\lim f(n)g(n)$, we can rewrite this as$$\lim\frac{f(n)}{{\tfrac{1}{g(n)}}}$$which is of the form $0/0$ or $\infty/\infty$ and so you can apply l'hopital's rule.
