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Greetings I am trying to solve $$\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$$ Using binomial series is pretty easy: $$\lim_{n\to\infty}n^2\left(1+\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)+1-\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-2\right)=\lim_{n\to\infty}n^2\left(-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)=-\frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?

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    $\begingroup$ rationalize twice. $\endgroup$ – Takahiro Waki Jul 8 '18 at 12:55
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    $\begingroup$ or $=n^2\left(\sqrt{1+\frac{1}{n}}-1-\frac{1}{2n}\right)+n^2\left(\sqrt{1-\frac{1}{n}}-1+\frac{1}{2n}\right)=-\frac{1}{4}\frac{1}{\sqrt{1+\frac{1}{n}}+1+\frac{1}{2n}}-\frac{1}{4}\frac{1}{\sqrt{1-\frac{1}{n}}-1+\frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge. $\endgroup$ – user566930 Jul 8 '18 at 13:03
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    $\begingroup$ @Dahaka LOL. What I was saying. Adding and subtracting $\frac{1}{2n}$ came from your Taylor expansion. $\endgroup$ – user566930 Jul 8 '18 at 13:07
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Hint: multiplying numerator and denominator by $\sqrt{1+1/n}+\sqrt{1-1/n}+2$ we get $$2 n^2 \frac{\sqrt{1-1/n^2}-1}{\sqrt{1+1/n}+\sqrt{1-1/n}+2}$$ and then do the same with $$\sqrt{1-1/n^2}+1$$ you will get

$$\frac{n^2(2(\sqrt{1-1/n^2}-1))(\sqrt{1-1/n^2}+1)}{(\sqrt{1+1/n}+\sqrt{1-1/n}+2)(\sqrt{1-1/n^2}+1)}$$

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    $\begingroup$ The most of this Problems can be solved by that way $\endgroup$ – Dr. Sonnhard Graubner Jul 8 '18 at 13:09
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Making $\delta = \frac 1n$ You can arrange it as

$$ \lim_{\delta\to 0}\left(\frac{\frac{\sqrt{1+\delta}-1}{\delta}-\frac{\sqrt{1-\delta}-1}{\delta}}{\delta}\right) = \left(\frac{d^2}{dx^2}\sqrt x\right)_{x=1} = -\frac 14 $$

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Here's a slightly different approach using multiplication by conjugates.

For notational simplicity, let $u=1/n$, so that $u\to0$ as $n\to\infty$. Then

$$n^2\left(\sqrt{1+{1\over n}}+\sqrt{1-{1\over n}}-2\right)={\sqrt{1+u}-1\over u^2}+{\sqrt{1-u}-1\over u^2}\\ ={1\over u(\sqrt{1+u}+1)}-{1\over u(\sqrt{1-u}+1)}\\={\sqrt{1-u}-\sqrt{1+u}\over u(\sqrt{1+u}+1)(\sqrt{1-u}+1)}\\={-2\over(\sqrt{1+u}+1)(\sqrt{1-u}+1)(\sqrt{1-u}+\sqrt{1+u})}\to{-2\over(1+1)(1+1)(1+1)}=-{1\over4}$$

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