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

Question

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?

Answer 1

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.

Answer 2

The last expression $$A=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$ leads to the result.

Since $n$ is large $$\sqrt{n+1}\sim \sqrt{n} \qquad \sqrt{n-1}\sim \sqrt{n}\qquad \sqrt{n^2-1}\sim \sqrt{n^2}=n$$

$$A \sim -\frac{2\sqrt{n^3}}{\big(\sqrt{n}+\sqrt{n}+2\sqrt{n}\big)\big( n+n\big)}=-\frac{2n\sqrt{n}}{\big(4\sqrt{n}\big)\big( 2n\big)}=-\frac 14$$

Answer 3

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 $$ -2 \cdot \frac {\sqrt n} {\sqrt{n+1}+\sqrt n} \cdot \frac{\sqrt n}{ \sqrt n + \sqrt{n-1}} \cdot \frac {\sqrt n}{\sqrt{n-1} + \sqrt{n+1}} \longrightarrow \frac{-1} 4. $$