Limit $\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$ 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?

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