<s>[INST] 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}}) $ 
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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!
 [/INST] $$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})=$$
$$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})\frac{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{({\sqrt{n+1}}+{\sqrt{n-1}})^2-4n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{2\sqrt{(n+1)(n-1)}+2n}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{4(n+1)(n-1)-4n^2}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{1}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty} n^{2/3}\frac{-4}{(\sqrt{(n+1)(n-1)}+2n)({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=$$
$$\lim_{n \to \infty} \frac{-4}{(\sqrt{(n^{1/3}+n^{-2/3})(n^{1/3}-n^{-2/3})}+2n^{1/3})({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=0$$
Misread $\frac{2}{3}$ as $\frac{3}{2}$ in the original post 
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