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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.