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### Let \(x_n := \frac{n^2-3}{n^2+1}\). Prove that \((x_n) \rightarrow 1\).

*Proof:* We compute that \(x_n =\frac{n^2-3}{n^2+1}=\frac{1-\tfrac{3}{n^2}}{1+\tfrac{1}{n^2}}\). Now
\(\left(1-\tfrac{3}{n^2}\right)\rightarrow 1\) because
\(p(x)=1-3x^2 \) is a polynomial and
\(\left(\tfrac{1}{n}\right)\rightarrow 0\). Similarly
\(\left(1+\tfrac{1}{n^2}\right)\rightarrow 1 \neq 0\) because
\(1+x^2\) is a polynomial. Thus
\(\left(\frac{1-\tfrac{3}{n^2}}{1+\tfrac{1}{n^2}}\right)\rightarrow \tfrac{1}{1}=1 \)
by CoS (Combinations of Sequences).