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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.