To solve the given limit problem, we need to find:
[tex]\[
\lim _{n \rightarrow \infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)
\][/tex]
First, let's simplify the expression inside the limit:
[tex]\[
\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2
\][/tex]
As [tex]\( n \to \infty \)[/tex], [tex]\( \frac{1}{n} \)[/tex] approaches 0. Therefore, we can expand the square roots using a Taylor series expansion around [tex]\( \frac{1}{n} = 0 \)[/tex].
For small [tex]\( x \)[/tex]:
[tex]\[
\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3)
\][/tex]
Using this expansion for [tex]\( x = \frac{1}{n} \)[/tex] and [tex]\( x = -\frac{1}{n} \)[/tex]:
1. [tex]\( \sqrt{1+\frac{1}{n}} \approx 1 + \frac{1}{2n} - \frac{1}{8n^2} + O(n^{-3}) \)[/tex]
2. [tex]\( \sqrt{1-\frac{1}{n}} \approx 1 - \frac{1}{2n} - \frac{1}{8n^2} + O(n^{-3}) \)[/tex]
Now, add these two expansions:
[tex]\[
\sqrt{1+\frac{1}{n}} + \sqrt{1-\frac{1}{n}} \approx \left(1 + \frac{1}{2n} - \frac{1}{8n^2}\right) + \left(1 - \frac{1}{2n} - \frac{1}{8n^2}\right)
\][/tex]
Combining like terms:
[tex]\[
\sqrt{1+\frac{1}{n}} + \sqrt{1-\frac{1}{n}} \approx 2 - \frac{1}{4n^2} + O(n^{-3})
\][/tex]
Subtract 2 from both sides:
[tex]\[
\sqrt{1+\frac{1}{n}} + \sqrt{1-\frac{1}{n}} - 2 \approx -\frac{1}{4n^2} + O(n^{-3})
\][/tex]
Now we multiply this by [tex]\( n^2 \)[/tex]:
[tex]\[
n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right) \approx n^2 \left(-\frac{1}{4n^2} + O(n^{-3})\right)
\][/tex]
Simplifying the product:
[tex]\[
n^2 \cdot -\frac{1}{4n^2} + n^2 \cdot O(n^{-3}) = -\frac{1}{4} + O(n^{-1})
\][/tex]
As [tex]\( n \to \infty \)[/tex], the higher-order term [tex]\( O(n^{-1}) \)[/tex] vanishes. Thus, we have:
[tex]\[
\lim_{n \rightarrow \infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right) = -\frac{1}{4}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{-\frac{1}{4}}
\][/tex]
