### Limits part 1: Infinite sequences: Techniques

### The squeeze lemma

Watch the animation of the sequence \(b_n = \frac{1+(-1)^n}{n}\) below:

The sequence consists of two parts: the constant sequence \(0\) and the sequence \(\frac{2}{n}\). These two sequences both converge to \(0\) as \(n\to\infty\), so we would like to conclude that \(\lim_{n\to\infty} b_n = 0\) as well. The squeeze lemma justifies this:

Squeeze lemma Suppose that \((a_n)_{n\in\mathbb N}, (b_n)_{n\in\mathbb N}\) and \((c_n)_{n\in\mathbb N}\) are three sequences with \(a_n \leq b_n \leq c_n\) for all \(n\). If \[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n,\] then \(b_n\) converges as well and the limit equals the limit of the other two sequences: \[ \lim_{n\to\infty} a_n =\lim_{n\to\infty} b_n= \lim_{n\to\infty} c_n\text.\]

In our example of \(b_n = \frac{1+(-1)^n}{n}\) we choose \(a_n = 0\) and \(c_n = \frac{2}{n}\) since \[0 \leq \frac{1+(-1)^n}{n} \leq \frac{2}{n}\] for all \(n\). The lemma now implies that \[ 0 \leq \lim_{n\to\infty} b_n \leq 0 \] So the limit equals \(0\).