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### Let \(x_n := \frac{n^2}{n^2+1}\). Prove that \((x_n) \rightarrow 1\).

*Proof:* Given \(\varepsilon >0\),
let \(K > \sqrt{\tfrac{1}{\varepsilon}-1}\), assuming that
\(\varepsilon < 1\).
Then if
\(n\ge K\) we have
\(|x_n - 1| = \)
\(\left|\frac{n^2}{n^2+1}-1\right|=\)
\(\left|\frac{-1}{n^2+1}\right|=\)
\(\frac{1}{n^2+1}\le\)
\(\frac{1}{K^2+1}< \)
\(\frac{1}{\left(\sqrt{\tfrac{1}{\varepsilon}-1}\right)^2+1}=\)
\(\varepsilon\).
Thus \(\left( \frac{n^2}{n^2+1}\right)\rightarrow 1\) by the Definition of Sequence Converence.