Learning Objectives

By the end of this course, you will be able to:

Prove and apply the uniqueness of limits theorem
Understand the relationship between convergence and boundedness
Apply the sign preservation property in proofs
Master arithmetic operations on convergent sequences
Understand the order preservation property and its limitations
Apply absolute value properties to limits

Section 1: Basic Properties

Theorem 2.1: Uniqueness of Limits

If a sequence converges, then its limit is unique.

Proof of Theorem 2.1:

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 $a = b$ .

∎

Theorem 2.2: Boundedness of Convergent Sequences

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

Proof of Theorem 2.2:

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 $n$ .

∎

Remark 2.1: Converse is False

The converse is FALSE. $\{(-1)^n\}$ is bounded (all terms between -1 and 1) but divergent because it oscillates forever.

Theorem 2.3: Sign Preservation

If $\lim_{n \to \infty} x_n = a > 0$ , then $\exists N$ such that for $n > N$ : $x_n > \frac{a}{2} > 0$ .

Proof of Theorem 2.3:

Take $\epsilon = \frac{a}{2} > 0$ . Then $\exists N$ : $n > N \Rightarrow |x_n - a| < \frac{a}{2}$

\Rightarrow a - \frac{a}{2} < x_n < a + \frac{a}{2} \Rightarrow x_n > \frac{a}{2} > 0

∎

Corollary 2.1: Non-negative Limit

If $x_n \geq 0$ for all $n$ and $\lim_{n \to \infty} x_n = a$ , then $a \geq 0$ .

Example 2.1: Sign Preservation Application

Problem: Show that if $\lim x_n = 3$ , then $x_n > 1$ for sufficiently large $n$ .

Solution:

By sign preservation with $a = 3 > 0$ , there exists $N$ such that for $n > N$ :

x_n > \frac{3}{2} = 1.5 > 1

Section 2: Arithmetic Operations

Theorem 2.4: Limit Laws

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

Operation	Result	Condition
Sum/Difference	$\lim_{n \to \infty} (x_n \pm y_n) = a \pm b$	None
Product	$\lim_{n \to \infty} (x_n \cdot y_n) = a \cdot b$	None
Scalar Multiple	$\lim_{n \to \infty} (c \cdot x_n) = c \cdot a$	c ∈ ℝ
Quotient	$\lim_{n \to \infty} \frac{x_n}{y_n} = \frac{a}{b}$	b ≠ 0, yₙ ≠ 0
Power	$\lim_{n \to \infty} x_n^k = a^k$	k ∈ ℕ

Proof of Product Rule:

We show $\lim(x_n y_n) = ab$ .

|x_n y_n - ab| = |x_n y_n - x_n b + x_n b - ab| \leq |x_n||y_n - b| + |b||x_n - a|

Since $\{x_n\}$ is convergent, it is bounded: $|x_n| \leq M$ for some $M > 0$ .

Given $\epsilon > 0$ , choose $N$ such that for $n > N$ :

$|x_n - a| < \frac{\epsilon}{2(|b|+1)}$
$|y_n - b| < \frac{\epsilon}{2M}$

Then $|x_n y_n - ab| < M \cdot \frac{\epsilon}{2M} + |b| \cdot \frac{\epsilon}{2(|b|+1)} < \epsilon$ .

∎

Remark 2.2: Indeterminate Forms

These laws require BOTH sequences to converge. For indeterminate forms like:

$\infty - \infty$
$0 \cdot \infty$
$\frac{0}{0}$ or $\frac{\infty}{\infty}$

Further analysis is needed (covered in CALC-2.3: Stolz theorem).

Example 2.2: Applying Limit Laws

Problem: 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 (sum, quotient, and $\lim \frac{1}{n^k} = 0$ ):

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

Example 2.3: Product of Bounded and Infinitesimal

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

Solution:

We have $|\sin n| \leq 1$ (bounded) and $\frac{1}{n} \to 0$ (infinitesimal).

\left|\frac{\sin n}{n}\right| = \frac{|\sin n|}{n} \leq \frac{1}{n} \to 0

By the domination principle: $\lim_{n \to \infty} \frac{\sin n}{n} = 0$ .

Example 2.4: Limit of Products

Problem: If $\lim x_n = 2$ and $\lim y_n = 3$ , find $\lim (x_n^2 - y_n^2)$ .

Solution:

Using the power rule and sum rule:

\lim (x_n^2 - y_n^2) = (\lim x_n)^2 - (\lim y_n)^2 = 2^2 - 3^2 = 4 - 9 = -5

Theorem 2.5: Quotient Rule (Detailed)

If $\lim x_n = a$ and $\lim y_n = b \neq 0$ , then $\lim \frac{x_n}{y_n} = \frac{a}{b}$ .

Proof of Theorem 2.5:

Step 1: By sign preservation, since $b \neq 0$ , there exists $N_1$ such that $|y_n| > \frac{|b|}{2}$ for $n > N_1$ .

Step 2: Estimate the difference:

\left|\frac{x_n}{y_n} - \frac{a}{b}\right| = \left|\frac{bx_n - ay_n}{by_n}\right| = \frac{|b(x_n - a) + a(b - y_n)|}{|b||y_n|}

Step 3: Using triangle inequality and $|y_n| > \frac{|b|}{2}$ :

\leq \frac{|b||x_n - a| + |a||y_n - b|}{|b| \cdot \frac{|b|}{2}} = \frac{2|x_n - a|}{|b|} + \frac{2|a||y_n - b|}{|b|^2}

Both terms can be made arbitrarily small for large $n$ .

∎

Corollary 2.2: Reciprocal Rule

If $\lim y_n = b \neq 0$ , then $\lim \frac{1}{y_n} = \frac{1}{b}$ .

Remark 2.3: Standard Limits to Remember

The following limits are fundamental and should be memorized:

\lim_{n \to \infty} \frac{1}{n^\alpha} = 0 \quad (\alpha > 0)

\lim_{n \to \infty} q^n = 0 \quad (|q| < 1)

\lim_{n \to \infty} \sqrt[n]{n} = 1

\lim_{n \to \infty} \sqrt[n]{a} = 1 \quad (a > 0)

Common Mistakes to Avoid

❌ Don't apply limit laws to divergent sequences

$\lim(n - n) \neq \lim n - \lim n$ (the right side is undefined)

❌ Don't divide when the denominator's limit is zero

The quotient rule requires $b \neq 0$

✓ Always verify convergence first

Before using any limit law, ensure all sequences involved converge

Section 3: Order Preservation

Theorem 2.5: Order Preservation

If $\{x_n\}$ and $\{y_n\}$ both converge, and $\exists N_0$ such that $x_n \leq y_n$ for all $n > N_0$ , then:

\lim_{n \to \infty} x_n \leq \lim_{n \to \infty} y_n

Critical Note: Strict Inequalities

Even if $x_n < y_n$ for ALL $n$ , the limits may be EQUAL.

Counterexample:

Let $x_n = 1 - \frac{1}{n}$ and $y_n = 1$ .

$x_n < y_n$ for all $n$
But $\lim x_n = 1 = \lim y_n$

Example 2.3: Order Preservation in Practice

Problem: If $0 < x_n < \frac{1}{n}$ for all $n$ , what is $\lim x_n$ ?

Solution:

We have $0 < x_n < \frac{1}{n}$ . Taking limits:

$\lim 0 = 0$
$\lim \frac{1}{n} = 0$

By order preservation: $0 \leq \lim x_n \leq 0$ , so $\lim x_n = 0$ .

(This is actually the Squeeze Theorem, covered in CALC-2.3)

Section 4: Absolute Value Property

Theorem 2.6: Limit Implies Absolute Value Limit

If $\lim_{n \to \infty} x_n = a$ , then $\lim_{n \to \infty} |x_n| = |a|$ .

Proof of Theorem 2.6:

By the reverse triangle inequality: $\big||x_n| - |a|\big| \leq |x_n - a|$

Given $\epsilon > 0$ , choose $N$ such that $|x_n - a| < \epsilon$ for $n > N$ .

Then $\big||x_n| - |a|\big| < \epsilon$ .

∎

Remark 2.3: Converse is False

The CONVERSE is FALSE!

Counterexample: $x_n = (-1)^n$

$|x_n| = 1 \to 1$ (converges)
$x_n$ diverges (oscillates between -1 and 1)

Theorem 2.7: Zero Limit Equivalence

\lim_{n \to \infty} x_n = 0 \Leftrightarrow \lim_{n \to \infty} |x_n| = 0

This equivalence holds ONLY for zero limits.

Proof of Theorem 2.7:

(⟹) Follows from Theorem 2.6 with $a = 0$ , giving $|a| = 0$ .

(⟸) If $|x_n| \to 0$ , then for any $\epsilon > 0$ , $\exists N$ : $n > N \Rightarrow |x_n| < \epsilon$ .

But $|x_n - 0| = |x_n|$ , so $x_n \to 0$ .

∎

Example 2.4: Using Absolute Value Properties

Problem: If $\lim x_n = -5$ , find $\lim |x_n|$ and $\lim x_n^2$ .

Solution:

By Theorem 2.6: $\lim |x_n| = |-5| = 5$

By the power rule (Theorem 2.4): $\lim x_n^2 = (-5)^2 = 25$

Section 5: Advanced Techniques

Theorem 2.8: Sandwich Principle for Products

If $\{x_n\}$ is bounded and $\lim_{n \to \infty} y_n = 0$ , then:

\lim_{n \to \infty} x_n y_n = 0

Proof of Theorem 2.8:

Since $\{x_n\}$ is bounded, $\exists M > 0$ such that $|x_n| \leq M$ for all $n$ .

Given $\epsilon > 0$ , since $y_n \to 0$ , $\exists N$ such that $|y_n| < \frac{\epsilon}{M}$ for $n > N$ .

Then $|x_n y_n| = |x_n||y_n| \leq M \cdot \frac{\epsilon}{M} = \epsilon$ .

∎

Example 2.5: Bounded Times Infinitesimal

Problem: Find $\lim_{n \to \infty} \frac{\cos n}{n}$ .

Solution:

We have $|\cos n| \leq 1$ (bounded) and $\frac{1}{n} \to 0$ (infinitesimal).

By Theorem 2.8: $\lim_{n \to \infty} \frac{\cos n}{n} = 0$ .

Theorem 2.9: Root Limit

If $\lim_{n \to \infty} x_n = a > 0$ and $k \in \mathbb{N}$ , then:

\lim_{n \to \infty} \sqrt[k]{x_n} = \sqrt[k]{a}

Proof of Theorem 2.9:

Let $f(x) = x^{1/k}$ for $x > 0$ . Using the factorization:

x^{1/k} - a^{1/k} = \frac{x - a}{x^{(k-1)/k} + x^{(k-2)/k}a^{1/k} + \cdots + a^{(k-1)/k}}

Since $x_n \to a > 0$ , the denominator is bounded away from 0 for large $n$ .

As $x_n - a \to 0$ , we get $x_n^{1/k} - a^{1/k} \to 0$ .

∎

Example 2.6: Combining Multiple Rules

Problem: Find $\lim_{n \to \infty} \sqrt{\frac{n^2 + 3n}{4n^2 + 1}}$ .

Solution:

First, find the limit inside the root:

\lim_{n \to \infty} \frac{n^2 + 3n}{4n^2 + 1} = \lim_{n \to \infty} \frac{1 + \frac{3}{n}}{4 + \frac{1}{n^2}} = \frac{1}{4}

By Theorem 2.9 (root limit):

\lim_{n \to \infty} \sqrt{\frac{n^2 + 3n}{4n^2 + 1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Example 2.7: Complex Expression

Problem: Find $\lim_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$ .

Solution:

Rationalize by multiplying by the conjugate:

\sqrt{n^2 + n} - n = \frac{(n^2 + n) - n^2}{\sqrt{n^2 + n} + n} = \frac{n}{\sqrt{n^2 + n} + n}

Divide by $n$ :

= \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \to \frac{1}{\sqrt{1} + 1} = \frac{1}{2}

Strategy Guide: When to Use Each Theorem

Rational Functions

Divide by highest power of $n$ , then apply sum/quotient rules

Expressions with Roots

Rationalize (multiply by conjugate), or use root limit theorem

Oscillating × Decaying

Use bounded × infinitesimal principle (Theorem 2.8)

Products and Quotients

Verify convergence first, then apply product/quotient rules

Remark 2.4: Hierarchy of Growth Rates

The following growth comparison is fundamental (from slowest to fastest):

\ln n \ll n^\alpha \ll a^n \ll n! \ll n^n \quad (\alpha > 0, a > 1)

Here $f(n) \ll g(n)$ means $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 0$ .

Example 2.8: Growth Rate Comparison

Problem: Find $\lim_{n \to \infty} \frac{n^{100}}{1.01^n}$ .

Solution:

By the growth hierarchy, polynomial $\ll$ exponential.

Since $1.01 > 1$ , the exponential $1.01^n$ dominates any polynomial $n^{100}$ .

Therefore: $\lim_{n \to \infty} \frac{n^{100}}{1.01^n} = 0$ .

Section 6: Worked Examples

Example 2.9: Limit of Recursive Sequence

Problem: Let $x_1 = 2$ , $x_{n+1} = \frac{x_n + 3}{2}$ . Find $\lim x_n$ if it exists.

Solution:

Step 1: Assume limit $L$ exists and solve $L = \frac{L + 3}{2}$ :

2L = L + 3 \Rightarrow L = 3

Step 2: Compute first few terms: $x_1 = 2, x_2 = 2.5, x_3 = 2.75, x_4 = 2.875$

The sequence is increasing and bounded above by 3.

Step 3: Prove boundedness: If $x_n < 3$ , then $x_{n+1} = \frac{x_n + 3}{2} < \frac{3 + 3}{2} = 3$ ✓

Step 4: Prove increasing: $x_{n+1} - x_n = \frac{x_n + 3}{2} - x_n = \frac{3 - x_n}{2} > 0$ (since $x_n < 3$ )

By Monotone Bounded Theorem, $x_n \to 3$ .

Example 2.10: Limit of Difference

Problem: Find $\lim_{n \to \infty} \left(\sqrt{n^2 + 2n} - \sqrt{n^2 - 2n}\right)$ .

Solution:

Rationalize:

\sqrt{n^2 + 2n} - \sqrt{n^2 - 2n} = \frac{(n^2 + 2n) - (n^2 - 2n)}{\sqrt{n^2 + 2n} + \sqrt{n^2 - 2n}} = \frac{4n}{\sqrt{n^2 + 2n} + \sqrt{n^2 - 2n}}

Divide by $n$ :

= \frac{4}{\sqrt{1 + \frac{2}{n}} + \sqrt{1 - \frac{2}{n}}} \to \frac{4}{1 + 1} = 2

Example 2.11: Product of Limits

Problem: If $\lim x_n = 2$ and $y_n = \frac{n+1}{n}$ , find $\lim x_n y_n$ .

Solution:

First, $\lim y_n = \lim \frac{n+1}{n} = \lim\left(1 + \frac{1}{n}\right) = 1$ .

By the product rule:

\lim x_n y_n = (\lim x_n)(\lim y_n) = 2 \cdot 1 = 2

Summary: Limit Properties Checklist

Arithmetic Rules

Sum: $\lim(x_n + y_n) = \lim x_n + \lim y_n$
Product: $\lim(x_n \cdot y_n) = \lim x_n \cdot \lim y_n$
Quotient: $\lim\frac{x_n}{y_n} = \frac{\lim x_n}{\lim y_n}$ (if denominator ≠ 0)
Power: $\lim x_n^k = (\lim x_n)^k$

Key Theorems

Uniqueness: limit is unique if exists
Convergent ⟹ Bounded
Sign Preservation
Order Preservation (non-strict)

Section 7: Proof Techniques

Theorem 2.10: Limit of Composite Sequences

If $\lim_{n \to \infty} x_n = a$ and $f$ is continuous at $a$ , then:

\lim_{n \to \infty} f(x_n) = f(a)

Example 2.12: Applying Composite Limit

Problem: Find $\lim_{n \to \infty} \sin\left(\frac{\pi n}{2n+1}\right)$ .

Solution:

First, $\lim_{n \to \infty} \frac{\pi n}{2n+1} = \lim_{n \to \infty} \frac{\pi}{2 + \frac{1}{n}} = \frac{\pi}{2}$

Since $\sin$ is continuous:

\lim_{n \to \infty} \sin\left(\frac{\pi n}{2n+1}\right) = \sin\left(\frac{\pi}{2}\right) = 1

Theorem 2.11: Dominated Convergence

If $|x_n| \leq y_n$ for all $n$ and $\lim y_n = 0$ , then $\lim x_n = 0$ .

Proof of Theorem 2.11:

Given $\epsilon > 0$ . Since $y_n \to 0$ , $\exists N$ such that $|y_n| < \epsilon$ for $n > N$ .

For $n > N$ : $|x_n - 0| = |x_n| \leq y_n = |y_n| < \epsilon$ .

∎

Example 2.13: Using Dominated Convergence

Problem: Find $\lim_{n \to \infty} \frac{(-1)^n \sin(n^2)}{n}$ .

Solution:

\left|\frac{(-1)^n \sin(n^2)}{n}\right| = \frac{|\sin(n^2)|}{n} \leq \frac{1}{n} \to 0

By dominated convergence: $\lim_{n \to \infty} \frac{(-1)^n \sin(n^2)}{n} = 0$ .

Remark 2.5: Telescoping Technique

For sequences involving differences, rewrite as telescoping sums:

x_n = x_1 + \sum_{k=1}^{n-1}(x_{k+1} - x_k)

This is useful when the differences have a pattern.

Example 2.14: Telescoping Sum

Problem: Find $\lim_{n \to \infty} \sum_{k=1}^{n}\frac{1}{k(k+1)}$ .

Solution:

Using partial fractions: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$

\sum_{k=1}^{n}\frac{1}{k(k+1)} = \sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1}

As $n \to \infty$ : $1 - \frac{1}{n+1} \to 1$ .

Proof Structure Template

State: "Let $\epsilon > 0$ be given."
Construct N: Define N explicitly in terms of ε.
Verify: Show that for $n > N$ , the inequality holds.
Conclude: "Therefore, $\lim x_n = a$ ."

Example 2.15: Complex Rational Limit

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

Solution:

Divide numerator and denominator by $n^3$ :

\frac{1 - \frac{2}{n} + \frac{1}{n^2}}{3 + \frac{1}{n^2} - \frac{1}{n^3}} \to \frac{1 - 0 + 0}{3 + 0 - 0} = \frac{1}{3}

Example 2.16: Nested Fractions

Problem: Find $\lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}$ .

Solution:

As $n \to \infty$ : $\frac{1}{n} \to 0$ , so $1 + \frac{1}{n} \to 1$ .

By quotient rule: $\frac{1}{1 + \frac{1}{n}} \to \frac{1}{1} = 1$ .

Theorem 2.12: Infinite Limit Laws

If $\lim x_n = +\infty$ and $\lim y_n = L > 0$ , then:

$\lim x_n \cdot y_n = +\infty$
$\lim (x_n + y_n) = +\infty$
$\lim \frac{y_n}{x_n} = 0$

Remark 2.6: Indeterminate Forms

The following forms require careful analysis:

\frac{0}{0}

\frac{\infty}{\infty}

0 \cdot \infty

\infty - \infty

1^\infty

0^0

Key Results Summary

Algebraic Properties

$\lim(x_n \pm y_n) = \lim x_n \pm \lim y_n$

$\lim(x_n \cdot y_n) = \lim x_n \cdot \lim y_n$

Order Properties

$x_n \leq y_n \Rightarrow \lim x_n \leq \lim y_n$

Convergent ⟹ Bounded

Example 2.17: Product with Zero Limit

Problem: Find $\lim_{n \to \infty} n \cdot \sin\frac{1}{n}$ .

Solution:

This is $\infty \cdot 0$ form. Rewrite as:

n \sin\frac{1}{n} = \frac{\sin\frac{1}{n}}{\frac{1}{n}}

Let $t = \frac{1}{n} \to 0$ . Then $\frac{\sin t}{t} \to 1$ .

Theorem 2.13: Ratio Comparison

If $\lim_{n \to \infty} \frac{x_n}{y_n} = L \neq 0$ and $y_n \to a$ , then $x_n \to La$ .

Remark 2.7: When Limit Laws Fail

Limit laws require all component limits to exist. Be careful with:

$\lim(x_n + y_n)$ when both diverge (may converge!)
$\lim(x_n \cdot y_n)$ with $0 \cdot \infty$ form
$\lim \frac{x_n}{y_n}$ with $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Example 2.18: Divergent Sum Converging

Problem: Let $x_n = n$ , $y_n = -n + \frac{1}{n}$ . Find $\lim(x_n + y_n)$ .

Solution:

Both $x_n \to +\infty$ and $y_n \to -\infty$ , but:

x_n + y_n = n + (-n + \frac{1}{n}) = \frac{1}{n} \to 0

Strategy Selection Guide

Polynomial ratio: Divide by highest power

With square roots: Rationalize (multiply by conjugate)

Exponential form: Use $e$ definition

Indeterminate: Algebraic manipulation first

Example 2.19: Difference of Square Roots

Problem: Find $\lim_{n \to \infty} (\sqrt{n^2 + n} - n)$ .

Solution:

Rationalize by multiplying by $\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$ :

\sqrt{n^2+n} - n = \frac{n^2+n-n^2}{\sqrt{n^2+n}+n} = \frac{n}{\sqrt{n^2+n}+n}

Divide by $n$ : $\frac{1}{\sqrt{1+\frac{1}{n}}+1} \to \frac{1}{1+1} = \frac{1}{2}$ .

Remark 2.8: Computing Tips

For $\sqrt{n^2 + an + b} - n$ , result is always $\frac{a}{2}$
For $\sqrt[3]{n^3 + an^2 + \ldots} - n$ , use binomial expansion
Factor out dominant terms before simplifying

Chapter 2.2 Mastery Checklist

Sum, difference, product rules
Quotient rule (denominator ≠ 0)
Order preservation (≤, not <)
Convergent ⟹ Bounded
Indeterminate form strategies

Practice Quiz: Limit Properties & Operations

8

Questions

0

Correct

0%

Accuracy

1

If

\lim x_n = 3

and

\lim y_n = 2

, what is

\lim(2x_n - 3y_n)

?

Easy

Not attempted

2

True or False: Every bounded sequence converges.

Easy

Not attempted

3

If

x_n > 0

for all

n

and

\lim x_n = L

, what can we conclude about

L

?

Medium

Not attempted

4

Which example shows that

x_n < y_n

for all

n

does NOT imply

\lim x_n < \lim y_n

?

Medium

Not attempted

5

If

\lim x_n = a \neq 0

, which statement is TRUE?

Medium

Not attempted

6

If

|x_n| \leq y_n

and

\lim y_n = 0

, what is

\lim x_n

?

Hard

Not attempted

7

If

\lim(x_n + y_n)

exists and

\lim x_n

exists, must

\lim y_n

exist?

Hard

Not attempted

8

What's wrong with: '

x_n < 1

for all

n

, so

\lim x_n < 1

'?

Hard

Not attempted

Frequently Asked Questions

Why does convergence imply boundedness but not vice versa?

Convergent sequences approach a specific value, so terms can't 'escape to infinity.' However, bounded sequences like {(-1)^n} can oscillate forever without settling down. Boundedness is necessary but not sufficient for convergence.

What is sign preservation and why is it useful?

If lim xₙ = a > 0, then xₙ > a/2 > 0 for large n. This lets us 'transfer' the positivity of the limit to the sequence terms. It's crucial for proving inequalities and for quotient rules (ensuring denominators stay away from zero).

Can the limit of a strict inequality be an equality?

Yes! If xₙ < yₙ for all n, we only get lim xₙ ≤ lim yₙ. Example: 1 - 1/n < 1 for all n, but lim(1 - 1/n) = 1 = lim 1. Strict inequalities become non-strict in the limit.

Why do limit laws require both sequences to converge?

For divergent sequences, expressions like ∞ - ∞ or 0 · ∞ are indeterminate. Example: xₙ = n and yₙ = n give xₙ - yₙ = 0 → 0, but xₙ = n and yₙ = n + 1 give xₙ - yₙ = -1 → -1. Same ∞ - ∞ form, different results.

What's the relationship between |xₙ| → |a| and xₙ → a?

xₙ → a implies |xₙ| → |a| (by reverse triangle inequality), but NOT vice versa. Exception: when a = 0, they're equivalent. Example: (-1)^n diverges but |(-1)^n| = 1 → 1.

Can I use limit laws before proving sequences converge?

No! You must first establish that both sequences converge (with finite limits). Only then can you apply the laws. A common error is writing 'lim(xₙ + yₙ) = lim xₙ + lim yₙ' when one of the limits doesn't exist.

Properties and Operations of Limits

Section 1: Basic Properties

Section 2: Arithmetic Operations

Section 3: Order Preservation

Section 4: Absolute Value Property

Section 5: Advanced Techniques

Rational Functions

Expressions with Roots

Oscillating × Decaying

Products and Quotients

Section 6: Worked Examples

Arithmetic Rules

Key Theorems

Section 7: Proof Techniques

Frequently Asked Questions