$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ 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?

Multiplying $\sqrt{n+1} - \sqrt n$ by $\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}$ yields $\dfrac 1 {\sqrt{n+1}+\sqrt n}.$
Similarly $\sqrt n - \sqrt{n-1} = \dfrac 1 {\sqrt n + \sqrt{n-1}}.$
So then we have
\begin{align}
& \big(\sqrt{n+1} - \sqrt n\big) - \big(\sqrt n - \sqrt{n-1} \big) \\[8pt]
= {} & \frac 1 {\sqrt{n+1}+\sqrt n} - \dfrac 1 {\sqrt n + \sqrt{n-1}} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \cdot \frac{\sqrt{n-1} + \sqrt{n+1}}{\sqrt{n-1} + \sqrt{n+1}} \\[12pt]
= {} & \frac{-2}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})(\sqrt{n-1} + \sqrt{n+1})}
\end{align}
If this is multiplied by $\sqrt{n^3}$ it becomes
$$
-2 \cdot \frac {\sqrt n} {\sqrt{n+1}+\sqrt n} \cdot \frac{\sqrt n}{ \sqrt n + \sqrt{n-1}} \cdot \frac {\sqrt n}{\sqrt{n-1} + \sqrt{n+1}} \longrightarrow \frac{-1} 4.
$$
