Section 5.3 Infinite limits
Definition 5.3.1.
Let \((x_{n})\) be a sequence of real numbers.
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 \(x_{n}\geq M\text{.}\)
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 \(x_{n}\leq M\text{.}\)
Example 5.3.2.
\(\sqrt{n}\rightarrow\infty\text{:}\) 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 \(n> N\text{,}\)
Thus, we have shown that for any \(M>0\) we can find \(N\) so that \(n> N\) implies \(\sqrt{n}\geq M\text{,}\) hence \(\sqrt{n}\rightarrow\infty\text{.}\)
Example 5.3.3.
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
Note that for \(n>1\text{,}\)
Thus, if \(N=M+1\text{,}\) then \(n> N\) implies