Sequences and Limits | Calculus Course

Section 1: Sequences and ε-N Definition

Definition 1.1: Sequence

A sequence is a function $f: \mathbb{N} \to \mathbb{R}$ . We denote it as $\{x_n\}_{n=1}^{\infty}$ or simply $\{x_n\}$ , where $x_n = f(n)$ .

Definition 1.2: Limit of a Sequence (ε-N Definition)

We say $\lim_{n \to \infty} x_n = a$ (or $x_n \to a$ ) if:

\forall \varepsilon > 0, \exists N \in \mathbb{N}: \forall n > N, |x_n - a| < \varepsilon

In words: For every positive tolerance ε, there exists a natural number N such that all terms after the Nth term are within ε of a.

Theorem 1.1: Uniqueness of Limits

If a sequence converges, then its limit is unique.

Proof of Theorem 1.1:

Suppose $\lim_{n \to \infty} x_n = a$ and $\lim_{n \to \infty} x_n = b$ with $a \neq b$ .

Let $\epsilon = \frac{|a-b|}{2} > 0$ .

By definition:

$\exists N_1$ : $n > N_1 \Rightarrow |x_n - a| < \epsilon$
$\exists N_2$ : $n > N_2 \Rightarrow |x_n - b| < \epsilon$

For $n > \max\{N_1, N_2\}$ :

|a - b| = |a - x_n + x_n - b| \leq |a - x_n| + |x_n - b| < 2\epsilon = |a - b|

This gives $|a - b| < |a - b|$ , a contradiction. Therefore $a = b$ .

∎

Example 1.1: Basic Limit

Prove: $\lim_{n \to \infty} \frac{1}{n} = 0$

Proof:

Given $\epsilon > 0$ , we need to find $N$ such that for $n > N$ :

\left|\frac{1}{n} - 0\right| = \frac{1}{n} < \epsilon

This is equivalent to $n > \frac{1}{\epsilon}$ .

Choose $N = \left\lfloor \frac{1}{\epsilon} \right\rfloor + 1$ .

For $n > N$ : $\frac{1}{n} < \frac{1}{N} < \epsilon$ . ∎

Example 1.2: Rational Function

Prove: $\lim_{n \to \infty} \frac{2n+1}{n+3} = 2$

Proof:

We compute:

\left|\frac{2n+1}{n+3} - 2\right| = \left|\frac{2n+1 - 2(n+3)}{n+3}\right| = \left|\frac{-5}{n+3}\right| = \frac{5}{n+3}

Given $\epsilon > 0$ , we need $\frac{5}{n+3} < \epsilon$ , i.e., $n > \frac{5}{\epsilon} - 3$ .

Choose $N = \max\left(0, \left\lfloor \frac{5}{\epsilon} - 3 \right\rfloor + 1\right)$ .

For $n > N$ : $\frac{5}{n+3} < \epsilon$ . ∎

Section 2: Limit Properties

Theorem 2.1: Boundedness of Convergent Sequences

If $\{x_n\}$ converges, then $\{x_n\}$ is bounded.

Proof of Theorem 2.1:

Let $\lim_{n \to \infty} x_n = L$ . Take $\epsilon = 1$ .

$\exists N$ : $n > N \Rightarrow |x_n - L| < 1 \Rightarrow |x_n| < |L| + 1$

Let $M = \max\{|x_1|, |x_2|, \ldots, |x_N|, |L| + 1\}$ .

Then $|x_n| \leq M$ for all $n$ .

∎

Theorem 2.2: Limit Laws

Let $\lim_{n \to \infty} x_n = a$ and $\lim_{n \to \infty} y_n = b$ . Then:

$\lim(x_n + y_n) = a + b$
$\lim(x_n \cdot y_n) = a \cdot b$
$\lim\frac{x_n}{y_n} = \frac{a}{b}$ (if $b \neq 0$ )
$\lim(cx_n) = ca$ for any constant $c$

Proof of Sum Rule:

We show $\lim(x_n + y_n) = a + b$ .

Given $\epsilon > 0$ :

$\exists N_1$ : $n > N_1 \Rightarrow |x_n - a| < \frac{\epsilon}{2}$
$\exists N_2$ : $n > N_2 \Rightarrow |y_n - b| < \frac{\epsilon}{2}$

Let $N = \max(N_1, N_2)$ . For $n > N$ :

|(x_n + y_n) - (a + b)| \leq |x_n - a| + |y_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

∎

Example 2.1: Applying Limit Laws

Find: $\lim_{n \to \infty} \frac{3n^2 + 2n - 1}{2n^2 - n + 5}$

Solution:

Divide numerator and denominator by $n^2$ :

\lim_{n \to \infty} \frac{3 + \frac{2}{n} - \frac{1}{n^2}}{2 - \frac{1}{n} + \frac{5}{n^2}}

Using limit laws:

= \frac{3 + 0 - 0}{2 - 0 + 0} = \frac{3}{2}

Section 3: Squeeze Theorem

Theorem 3.1: Squeeze Theorem (Sandwich Theorem)

Let $\{a_n\}$ , $\{b_n\}$ , $\{c_n\}$ be sequences satisfying:

$\exists N_0 \in \mathbb{N}$ : $a_n \leq b_n \leq c_n$ for all $n \geq N_0$
$\lim_{n \to \infty} a_n = L = \lim_{n \to \infty} c_n$

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

Proof of Theorem 3.1:

Given $\epsilon > 0$ :

$\exists N_1$ : $n > N_1 \Rightarrow |a_n - L| < \epsilon \Rightarrow L - \epsilon < a_n$
$\exists N_2$ : $n > N_2 \Rightarrow |c_n - L| < \epsilon \Rightarrow c_n < L + \epsilon$

For $n > \max\{N_0, N_1, N_2\}$ :

L - \epsilon < a_n \leq b_n \leq c_n < L + \epsilon

Therefore $|b_n - L| < \epsilon$ .

∎

Example 3.1: Using Squeeze Theorem

Find: $\lim_{n \to \infty} \frac{\sin n}{n}$

Solution:

We have $-1 \leq \sin n \leq 1$ , so:

-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}

Since $-\frac{1}{n} \to 0$ and $\frac{1}{n} \to 0$ , by Squeeze Theorem:

\lim_{n \to \infty} \frac{\sin n}{n} = 0

Section 4: Monotone Bounded Theorem

Definition 4.1: Monotone Sequence

A sequence $\{a_n\}$ is:

Monotonically increasing if $a_n \leq a_{n+1}$ for all $n$
Monotonically decreasing if $a_n \geq a_{n+1}$ for all $n$

Theorem 4.1: Monotone Bounded Theorem

A bounded monotone sequence converges.

Proof of Theorem 4.1 (Increasing Case):

Let $E = \{a_n : n \in \mathbb{N}\}$ . Since $E$ is non-empty and bounded above, by the Completeness Axiom, $\alpha = \sup E$ exists.

We claim $\lim_{n \to \infty} a_n = \alpha$ .

Given $\epsilon > 0$ :

By supremum property: $\exists N$ such that $a_N > \alpha - \epsilon$
By monotonicity: $n > N \Rightarrow a_n \geq a_N > \alpha - \epsilon$
By upper bound: $a_n \leq \alpha < \alpha + \epsilon$

Therefore $|a_n - \alpha| < \epsilon$ for $n > N$ .

∎

Example 4.1: The Number e

Prove: The sequence $x_n = (1 + \frac{1}{n})^n$ converges.

Solution:

One can show that $\{x_n\}$ is strictly increasing and bounded above by 3.

By the Monotone Bounded Theorem, it converges. The limit is denoted $e$ :

e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828...

Section 5: Subsequences and Bolzano-Weierstrass

Definition 5.1: Subsequence

Let $\{a_n\}$ be a sequence. If $\{n_k\} \subset \mathbb{N}$ with $n_1 < n_2 < n_3 < \cdots$ , then $\{a_{n_k}\}$ is a subsequence of $\{a_n\}$ .

Theorem 5.1: Subsequence Convergence

$\{a_n\}$ converges to $a$ $\Leftrightarrow$ Every subsequence of $\{a_n\}$ converges to $a$ .

Theorem 5.2: Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Proof of Theorem 5.2:

Method: Every sequence has a monotone subsequence (by peak argument). Since the original is bounded, the monotone subsequence is bounded, hence converges by Monotone Bounded Theorem.

∎

Example 5.1: Divergence via Subsequences

Prove: $\{(-1)^n\}$ diverges.

$\{a_{2k}\} = 1, 1, 1, \ldots \to 1$

$\{a_{2k-1}\} = -1, -1, -1, \ldots \to -1$

Since $1 \neq -1$ , the sequence diverges.

Section 6: Cauchy Sequences

Definition 6.1: Cauchy Sequence

A sequence $\{a_n\}$ is a Cauchy sequence if:

\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n, m > N: |a_n - a_m| < \epsilon

Theorem 6.1: Cauchy Criterion

A sequence of real numbers converges if and only if it is a Cauchy sequence.

Proof of Theorem 6.1 (Necessity):

Let $\lim a_n = L$ . Given $\epsilon > 0$ , $\exists N$ : $n > N \Rightarrow |a_n - L| < \frac{\epsilon}{2}$ .

For $n, m > N$ :

|a_n - a_m| \leq |a_n - L| + |L - a_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

∎

Example 6.1: Harmonic Series is NOT Cauchy

Show: $H_n = \sum_{k=1}^n \frac{1}{k}$ is not Cauchy.

Solution:

|H_{2n} - H_n| = \sum_{k=n+1}^{2n} \frac{1}{k} > n \cdot \frac{1}{2n} = \frac{1}{2}

This violates the Cauchy criterion, so the harmonic series diverges.

Section 7: Nested Interval Theorem

Theorem 7.1: Cantor's Nested Interval Theorem

If $\{[a_n, b_n]\}$ is a sequence of closed intervals with:

$[a_n, b_n] \supseteq [a_{n+1}, b_{n+1}]$ for all $n$
$\lim_{n \to \infty} (b_n - a_n) = 0$

Then $\exists!\, \xi \in \mathbb{R}$ : $\{\xi\} = \bigcap_{n=1}^{\infty} [a_n, b_n]$

Proof of Theorem 7.1:

The sequence $\{a_n\}$ is increasing and bounded above by $b_1$ , so it converges to some $\xi$ .

Similarly, $\{b_n\}$ is decreasing and bounded below, converging to the same $\xi$ (since $b_n - a_n \to 0$ ).

Therefore $\xi \in [a_n, b_n]$ for all $n$ , and is unique.

∎

Example 7.1: Bisection Method

Application: The bisection method for finding roots uses nested intervals. Each step halves the interval, and the intersection gives the root.

Work These Ideas by Hand — Then Check Yourself

Number Sequence Calculator — identify patterns and limits of arithmetic and geometric sequences
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Practice Quiz: Sequences and Limits

10

Questions

0

Correct

0%

Accuracy

1

Which of the following sequences is convergent?

Easy

Not attempted

2

The ε-N definition states:

\lim_{n \to \infty} x_n = a

if and only if...

Easy

Not attempted

3

If

\lim x_n = a

and

\lim y_n = b

, then

\lim(x_n + y_n)

equals:

Easy

Not attempted

4

The Squeeze Theorem applies when:

Medium

Not attempted

5

A bounded monotone sequence:

Medium

Not attempted

6

If

|x_n| \leq y_n

and

\lim y_n = 0

, what is

\lim x_n

?

Medium

Not attempted

7

The sequence

x_n = (1 + \frac{1}{n})^n

converges to:

Hard

Not attempted

8

If

\lim(x_n + y_n)

exists and

\lim x_n

exists, must

\lim y_n

exist?

Hard

Not attempted

9

What's wrong with: '

x_n < 1

for all n, so

\lim x_n < 1

'?

Hard

Not attempted

10

The Stolz Theorem is useful for:

Hard

Not attempted

Frequently Asked Questions

Why do we need the ε-N definition?

The ε-N definition provides a rigorous framework for limits, eliminating ambiguity. It allows us to prove statements about limits with certainty and handle edge cases that intuitive approaches miss.

What's the intuition behind ε-N?

Think of ε as a tolerance level - how close we want the sequence to be to the limit. N is the starting point after which all terms are within this tolerance. The definition says: no matter how tight we make the tolerance, we can always find a point after which all terms satisfy it.

Can a sequence have more than one limit?

No! If a sequence converges, its limit is unique. This is a fundamental theorem that we prove using the ε-N definition.

What's the difference between bounded and convergent?

Convergent implies bounded, but the converse is false. The sequence {(-1)^n} is bounded but divergent because it oscillates. However, every convergent sequence must be bounded.

How do I choose the right N in a proof?

Work backwards: start with |xₙ - a| < ε and algebraically solve for what condition on n makes this true. The resulting expression tells you how to choose N, often using floor functions or adding 1 to ensure N is a natural number.