The Squeeze Theorem for Convergent Sequences Examples 1

# The Squeeze Theorem for Convergent Sequences Examples 1

Recall from The Squeeze Theorem for Convergent Sequences page that if the limits of the sequences $\{ a_n \}$, $\{ b_n \}$, and $\{ c_n \}$ are convergent and $a_n ≤ b_n ≤ c_n$ is true always after some $n^{\mathrm{th}}$ term, then if $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$, then $\lim_{n \to \infty} b_n = L$

We will now look at some examples of applying the Squeeze theorem.

## Example 1

Determine the limit of the sequence $\left \{ \frac{\cos n}{n} \right \}$.

First note that $-1 ≤ \cos n ≤ 1$ for all $n \in \mathbb{N}$. Therefore we have that:

(1)
\begin{align} \quad -\frac{1}{n} ≤ \frac{\cos n}{n} ≤ \frac{1}{n} \end{align}

Now $\lim_{n \to \infty} \frac{1}{n} = 0$ and $\lim_{n \to \infty} -\frac{1}{n} = 0$, and so by the Squeeze theorem we have that $\lim_{n \to \infty} \frac{\cos n}{n} = 0$.

## Example 2

Determine the limit of the sequence $\left \{ \frac{\sin n + \cos n}{n^2} \right \}$

Once again we note that $-1 ≤ \sin n ≤ 1$ and $-1 ≤ \cos n ≤ 1$ so $-2 ≤ \sin n + \cos n ≤ 2$ and thus:

(2)
\begin{align} \quad - \frac{2}{n^2} ≤ \frac{\sin n + \cos n}{n^2} ≤ \frac{2}{n^2} \end{align}

Now $\lim_{n \to \infty} -\frac{2}{n^2} = 0$ and $\lim_{n \to \infty} \frac{2}{n^2} = 0$ so by the Squeeze theorem we have that $\lim_{n \to \infty} \frac{\sin n + \cos n}{n^2} = 0$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License