$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$
That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$
If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
Now that doesn't seem to be of any use either. Any ideas?

Dividing both the numerator and the denominator by $\sqrt{n^3}$, you have
\begin{eqnarray}
&&\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}\\
&=&\frac{2}{\big(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}+2\big)\big( \sqrt{1-\frac1{n^2}}+1\big)}.
\end{eqnarray}
Now you can take the limit to get the result.
