[pdf]

The Limit Theorems (Part II)

MATH 4630 · Chapter 2: Sequences and Series

We continue our exploration of how limits interact with properties of sequences. This section addresses order relationships and the powerful Squeeze Theorem, which allows us to determine limits by comparison.

Order Limit Theorem

Theorem: Order Limit Theorem

Suppose $(a_n) \to a$, $(b_n) \to b$, and $a_n \leq b_n$ for all $n \in \mathbb{N}$. Then

$$a \leq b$$

Important Note

Notice that we have $a_n \leq b_n$ (strict inequality is not always preserved), but we conclude $a \leq b$ (not $a < b$). This is a critical distinction.

Proof (Fill in the steps)

Setup:

We'll prove by contradiction. Suppose $a > b$. Then $a - b > 0$.

Step 1: Apply definition of convergence

Let $\varepsilon = \dfrac{a-b}{2}$. Since $(a_n) \to a$, there exists $N_1 \in \mathbb{N}$ such that $n \geq N_1$ implies

$$|a_n - a| < $$

Since $(b_n) \to b$, there exists $N_2 \in \mathbb{N}$ such that $n \geq N_2$ implies

$$|b_n - b| < $$

Step 2: Combine inequalities

Let $N = \max\{N_1, N_2\}$ and pick $n \geq N$. Show that $a_n > b_n$, which contradicts the hypothesis.

Conclusion:

Since $a > b$ leads to a contradiction, we must have $a \leq b$. $\square$

The Squeeze Theorem

Theorem: Squeeze Theorem (Sandwich Theorem)

Suppose $(a_n)$, $(b_n)$, and $(c_n)$ are three sequences such that:

  1. $a_n \leq b_n \leq c_n$ for all $n \in \mathbb{N}$, and
  2. $\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} c_n = L$.

Then $\lim\limits_{n \to \infty} b_n = L$.

0 $n$ $L$ $(c_n)$ $(b_n)$ $(a_n)$

Proof of Squeeze Theorem

Setup:

Let $\varepsilon > 0$ be arbitrary. Since $\lim\limits_{n \to \infty} a_n = L$, there exists $N_1 \in \mathbb{N}$ such that

$$|a_n - L| < \varepsilon \quad \text{whenever} \quad n \geq N_1$$

This means

$$L - \varepsilon < a_n < L + \varepsilon$$

Similarly, since $\lim\limits_{n \to \infty} c_n = L$, there exists $N_2 \in \mathbb{N}$ such that

$$L - \varepsilon < c_n < L + \varepsilon \quad \text{whenever} \quad n \geq N_2$$

Combining the inequalities:

Let $N = \max\{N_1, N_2\}$. For $n \geq N$:

Therefore, $\lim\limits_{n \to \infty} b_n = L$. $\square$

Examples

Example 1: The Squeeze Theorem in Action

Find $\lim\limits_{n \to \infty} \dfrac{\sin(n)}{n}$.

Solution:

We know that $-1 \leq \sin(n) \leq 1$ for all $n$. Dividing by $n > 0$:

$$ \leq \frac{\sin(n)}{n} \leq $$

Now, $\lim\limits_{n \to \infty} \left(\right) = 0$ and $\lim\limits_{n \to \infty} \left(\right) = 0$.

By the Squeeze Theorem:

$$\lim_{n \to \infty} \frac{\sin(n)}{n} = $$

Example 2: A Bounded Term Sequence

Find $\lim\limits_{n \to \infty} \dfrac{(-1)^n}{n}$.

Solution:

Observe that for all $n \in \mathbb{N}$:

$$-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}$$

We have $\lim\limits_{n \to \infty} \left(-\dfrac{1}{n}\right) = $ and $\lim\limits_{n \to \infty} \left(\dfrac{1}{n}\right) = $ .

By the Squeeze Theorem:

$$\lim_{n \to \infty} \frac{(-1)^n}{n} = $$
Exercises

Exercise 1

For each $n$, let $a_n = \dfrac{n + (-1)^n}{n}$. Prove that $\lim\limits_{n \to \infty} a_n = 1$ using the Squeeze Theorem.

Hint: Rewrite to isolate the bounded term.

Exercise 2

Prove that $\lim\limits_{n \to \infty} \dfrac{\cos(n^2)}{n} = 0$ using the Squeeze Theorem.

Key Takeaway

The Squeeze Theorem is invaluable when you cannot compute a limit directly but can bound a sequence between two sequences whose limits you know.

← Back to MATH 4630