Find the limit of the sequence section 3.2.6
Consider the sequence, n−1nlimn→∞n−1n=limn→∞(1−1n)=limn→∞1−limn→∞1nWhere limn→∞1n=0Hence,limn→∞n−1n=1\displaystyle \text{Consider the sequence, $\frac{n-1}{n}$}\\ \lim_{n \to \infty}\frac{n-1}{n} = \lim_{n \to \infty} (1-\frac1n)\\ =\lim_{n \to \infty}1 - \lim_{n \to \infty} \frac1n\\ \text{Where $\lim_{n \to \infty} \frac1n = 0$}\\ \text{Hence,} \lim_{n \to \infty}\frac{n-1}{n}=1Consider the sequence, nn−1n→∞limnn−1=n→∞lim(1−n1)=n→∞lim1−n→∞limn1Where limn→∞n1=0Hence,n→∞limnn−1=1
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