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Section 5.3 Infinite limits

Definition 5.3.1.

Let \((x_{n})\) be a sequence of real numbers.

  1. We say \((x_{n})\) tends to \(+\infty\) or diverges to \(\infty\) or \(x_{n}\rightarrow \infty\) as \(n \to \infty\) if, for every \(M>0\text{,}\) there is \(N\) so that \(n> N\) implies \(x_{n}\geq M\text{.}\)

  2. Similarly, we say \(x_{n}\) tends or diverges to \(-\infty\) or \(x_{n}\rightarrow -\infty\) as \(n \to \infty\) if, for all \(M\lt 0\text{,}\) there is \(N\) so that \(n>N\) implies \(x_{n}\leq M\text{.}\)

\(\sqrt{n}\rightarrow\infty\text{:}\) let \(M>0\text{,}\) we need to find \(N\) so that \(n> N\) implies \(\sqrt{n}> M\text{.}\) If we pick \(N=M^2\text{,}\) then for \(n> N\text{,}\)

\begin{equation*} \sqrt{n}> \sqrt{N}=\sqrt{M^2}=M\text{.} \end{equation*}

Thus, we have shown that for any \(M>0\) we can find \(N\) so that \(n> N\) implies \(\sqrt{n}\geq M\text{,}\) hence \(\sqrt{n}\rightarrow\infty\text{.}\)

The sequence \(x_{n} = n^2-n-1\) tends to infinity. To prove this, we must show that there is \(N\) so that \(n> N\) implies

\begin{equation*} n^2-n-1>M\text{,} \end{equation*}

Note that for \(n>1\text{,}\)

\begin{equation*} n^2-n -1= n(n-1)-1\geq n\cdot 1-1=n-1 \end{equation*}

Thus, if \(N=M+1\text{,}\) then \(n> N\) implies

\begin{equation*} n^2-n-1 \geq n-1> N-1=M+1-1=M\text{.} \end{equation*}