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Limits & Continuity (AP Calculus AB Unit 1)

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Limits & Continuity (AP Calculus AB Unit 1) - Complete Interactive Lesson

Part 1: The Foundation of Calculus

∫ Understanding Limits

Part 1 of 7 — The Foundation of Calculus

Topics in This Part

Section
📖 What Is a Limit?
Direct Substitution
📌 The Indeterminate Form $\frac{0}{0}$
Algebraic Techniques: Factor, Rationalize, Expand
When Limits Do Not Exist

🔑 Key Concept: A limit describes the value a function approaches as the input gets closer to a particular value. The function does NOT need to be defined at that point for the limit to exist.

📖 What Is a Limit?

A limit describes the value a function approaches as $x$ approaches a particular value $c$ :

$\boxed{\lim_{x \to c} f(x) = L}$

📖 Evaluating Limits by Direct Substitution

The simplest method: just plug in the value. If $f(c)$ produces a real number, then:

$\boxed{\lim_{x \to c} f(x) = f(c) \quad \text{(if } f \text{ is continuous at } c\text{)}}$

Check Your Understanding 🎯

📌 The Indeterminate Form $\frac{0}{0}$

When direct substitution gives $\frac{0}{0}$ , you have an . The limit may still exist — you must apply algebraic techniques to simplify.

More Practice 🎯

When Limits Do Not Exist (DNE)

A limit does not exist when:

Situation	Example	Why DNE
Left ≠ Right	$\lim_{x \to 0} \frac{	x
Unbounded	$\lim_{x \to 0} \frac{1}{x^2}$

Key Techniques Summary

Situation	Strategy	Result Example
Direct sub works	Plug in $c$	$\lim_{x \to 2} x^3 = 8$

Match the Technique 🔍

For each limit, select the best first step.

Compute the Limit ✍️

Part 2: Mastering Limit Computation

∫ Evaluating Limits Algebraically

Part 2 of 7 — Mastering Limit Computation

Topics in This Part

Section
📖 Special Trig Limits
Limits at Infinity for Rational Functions
📌 Limits Involving $e$
Piecewise Function Limits
One-Sided Limits

🔑 Key Concept: Beyond factoring and rationalizing, certain memorized limits and comparison strategies let you evaluate limits quickly on the AP exam.

📖 Special Trig Limits

Two limits you must memorize for the AP exam:

$\lim_{x \to 0} \frac{\sin x}{x} = 1 \underset{}{}$

Part 3: Left-Hand and Right-Hand Limits

∫ One-Sided Limits

Part 3 of 7 — Left-Hand and Right-Hand Limits

Topics in This Part

Section
📖 Definition of One-Sided Limits
Piecewise Functions & Breakpoints
📌 Vertical Asymptotes & One-Sided Behavior
Absolute Value Functions
The Two-Sided Limit Existence Theorem

🔑 Key Concept: A two-sided limit exists if and only if both one-sided limits exist and are equal. Mastering one-sided limits is essential for analyzing piecewise functions and asymptotic behavior.

📖 Left-Hand and Right-Hand Limits

The left-hand limit approaches $c$ from values less than $c$ :

Part 4: Bounding Limits

∫ The Squeeze Theorem

Part 4 of 7 — Bounding Limits

Topics in This Part

Section
📖 Statement of the Squeeze Theorem
Classic Oscillation Examples
📌 Proving $\frac{\sin x}{x} \to 1$
When to Use (and When Not To)

Part 5: When Functions Behave Nicely

∫ Continuity & the Intermediate Value Theorem

Part 5 of 7 — When Functions Behave Nicely

Topics in This Part

Section
📖 The Three Conditions for Continuity
Types of Discontinuities
📌 Continuity on an Interval
Functions That Are Always Continuous
The Intermediate Value Theorem (IVT)

🔑 Key Concept: A function is continuous at a point when its limit equals its function value. The IVT guarantees that continuous functions on closed intervals take on every intermediate value — a powerful existence theorem.

📖 The Three Conditions for Continuity at $x = c$

$f is continuous at x = c$

Part 6: AP-Level Practice

∫ Problem-Solving Workshop

Part 6 of 7 — AP-Level Practice

Strategy Decision Tree

Step	Action	If Result Is...
1	Try direct substitution	A number → done!
2a	Got $\frac{0}{0}$ ?	Factor, rationalize, or use trig identities
2b	Got ?

Part 7: Putting It All Together

∫ Review & AP Exam Applications

Part 7 of 7 — Putting It All Together

Complete Limits & Continuity Toolkit

Tool	When to Use	Key Formula
Direct Substitution	Always try first	Plug in $x = c$
Factoring	$\frac{0}{0}$ with polynomials

This means: as $x$ gets arbitrarily close to $c$ (from both sides), $f(x)$ gets arbitrarily close to $L$ .

Statement	What It Means
$\lim_{x \to c} f(x) = L$	$f(x)$ approaches $L$ as $x$ approaches $c$
$f(c) = L$	The function equals $L$ at $x = c$

These are different things! A function can have a limit at a point where it's not defined, or where $f(c)$ differs from $L$ .

Consider a function with a hole at $(3, 7)$ and $f(3) = 2$ .

$\lim_{x \to 3} f(x) = 7$ (the function heads toward 7)
$f(3) = 2$ (the actual function value is 2)

AP Tip: About 15% of AP Calculus MC questions involve limits. Mastering this concept is foundational for derivatives and integrals.

Functions Where Direct Substitution Always Works

Function Type	Example
Polynomials	$\lim_{x \to 3} (2x^2 + 1) = 19$
Exponentials	$\lim_{x \to 0} e^{2x} = 1$
Trig functions	$\lim_{x \to \pi} \sin(x) = 0$
Rational (if denominator $\neq 0$ )	$\lim_{x \to 1} \frac{x+3}{x+1} = 2$

$\lim_{x \to 2} (x^3 - 4x + 7)$

Substitute $x = 2$ : $8 - 8 + 7 = 7$ ✓

🔑 Key Fact: All polynomial and exponential functions are continuous everywhere, so direct substitution always works for them.

indeterminate form

$\boxed{\frac{0}{0} \text{ means "do more algebra" — NOT "does not exist"}}$

Technique 1: Factoring

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4$

Technique 2: Rationalizing (Conjugate Multiplication)

$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$

Multiply numerator and denominator by the conjugate $\sqrt{x+4} + 2$ :

$= \frac{(x+4) - 4}{x(\sqrt{x+4}+2)} = \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2}$

At $x = 0$ : $\frac{1}{2+2} = \frac{1}{4}$

Technique 3: Expanding

$\lim_{x \to 0} \frac{(x+3)^2 - 9}{x} = \lim_{x \to 0} \frac{x^2+6x+9-9}{x} = \lim_{x \to 0} \frac{x^2+6x}{x} = \lim_{x \to 0}(x+6) = 6$

AP Tip: On free-response questions, always show the algebraic simplification step. Simply writing the final answer without work earns 0 points.

$\frac{\text{nonzero}}{0}$ — Not Indeterminate!

$\frac{\text{nonzero}}{0} \Rightarrow \text{Check } \pm\infty \text{ or DNE}$

Example: $\lim_{x \to 3} \frac{1}{(x-3)^2} = +\infty$ (both sides go to $+\infty$ )

Example: $\lim_{x \to 3} \frac{1}{x-3}$ → DNE (left goes to $-\infty$ , right goes to $+\infty$ )

🔑 Key Distinction: $\frac{0}{0}$ = indeterminate (do more work). $\frac{\text{nonzero}}{0}$ = usually $\pm\infty$ or DNE.

AP Tip: On the AP exam, the answer choice "does not exist" is tempting but usually wrong for $\frac{0}{0}$ forms. Always try algebra first!

x \to 0 lim \frac{sin x}{x} = 1 x \to 0 lim \frac{1 - cos x}{x} = 0

Extending the Pattern

The key insight: you can match the argument of $\sin$ with the denominator.

Limit	Rewriting	Result
$\frac{\sin(3x)}{x}$	$3 \cdot \frac{\sin(3x)}{3x}$	$3$
$\frac{\sin(ax)}{bx}$	$\frac{a}{b} \cdot \frac{\sin(ax)}{ax}$
$\frac{\sin(5x)}{\sin(2x)}$	$\frac{5 x}{2 x} \cdot \frac{\sin (5 x) /}{}$
$\frac{\tan x}{x}$	$\frac{\sin x}{x} \cdot \frac{1}{\cos x}$

$\boxed{\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}}$

$\lim_{x \to 0} \frac{\sin(7x)}{\sin(3x)} = \lim_{x \to 0} \frac{\sin(7x)}{7x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{7}{3} = 1 \cdot 1 \cdot \frac{7}{3} = \frac{7}{3}$

AP Tip: These trig limits appear in disguised forms nearly every year. The secret is always to make the argument of $\sin$ match the denominator.

Check Your Understanding 🎯

📖 Limits at Infinity for Rational Functions

For rational functions $\frac{P(x)}{Q(x)}$ as $x \to \pm\infty$ , compare the degrees:

$\boxed{\lim_{x \to \infty} \frac{a_n x^n + \cdots}{b_m x^m + \cdots} = \begin{cases} 0 & \text{if } n < m \\ \frac{a_n}{b_m} & \text{if } n = m \\ \pm\infty & \text{if } n > m \end{cases}}$

Memory Aid

Degree Comparison	Result	Mnemonic
deg(top) < deg(bottom)	$0$	"Bottom Heavy → squishes to 0"
deg(top) = deg(bottom)	$\frac{\text{leading coeff}}{\text{leading coeff}}$	"Tie → compare captains"
deg(top) > deg(bottom)

Worked Examples

Example 1: $\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - 2} = \frac{3}{5}$ (same degree: ratio of leading coefficients)

Example 2: $\lim_{x \to \infty} \frac{2x}{x^2 + 1} = 0$ (degree 1 < degree 2: bottom wins)

Example 3: $\lim_{x \to \infty} \frac{x^3}{x+1} = \infty$ (degree 3 > degree 1: top wins)

🔑 Key Fact: Horizontal asymptotes come directly from limits at infinity. If $\lim_{x \to \infty} f(x) = L$ , then $y = L$ is a horizontal asymptote.

Limits at Infinity Practice 🎯

📌 Limits Involving $e$

The number $e$ is defined by:

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

Important Variants

Limit	Result	When You See It
$\lim_{x \to 0} \frac{e^x - 1}{x}$

AP Tip: The limit $\frac{e^x - 1}{x} \to 1$ is really $f'(0)$ where . Recognizing limits as derivatives in disguise is a powerful exam strategy.

Piecewise Function Limits & One-Sided Limits

For piecewise functions, evaluate the limit from each side separately:

$f(x) = \begin{cases} x^2 & x < 1 \\ 2x - 1 & x \geq 1 \end{cases}$

$\lim_{x \to 1^-} f(x) = 1^2 = 1$ (use the "" rule)

Since both sides agree: $\lim_{x \to 1} f(x) = 1$ ✓

When a Piecewise Limit Fails

$g(x) = \begin{cases} x + 2 & x < 3 \\ x^2 - 4 & x \geq 3 \end{cases}$

$\lim_{x \to 3^-} g(x) = 5$
$\lim_{x \to 3^+} g(x) = 5$

These agree, so $\lim_{x \to 3} g(x) = 5$ . But $g(3) = 5$ too, so is also continuous here.

$h(x) = \begin{cases} 2x & x < 1 \\ x + 3 & x \geq 1 \end{cases}$

$\lim_{x \to 1^-} h(x) = 2$
$\lim_{x \to 1^+} h(x) = 4$

Left $\neq$ right, so $\lim_{x \to 1} h(x)$ does not exist.

🔑 Key Fact: A two-sided limit exists if and only if both one-sided limits exist and are equal.

Evaluate Each Limit 🔍

Compute the Limit ✍️

x \to c^{-} lim f (x) = L_{1}

The right-hand limit approaches $c$ from values greater than $c$ :

$\boxed{\lim_{x \to c^+} f(x) = L_2}$

The Existence Theorem

$\boxed{\lim_{x \to c} f(x) = L \quad \Longleftrightarrow \quad \lim_{x \to c^-} f(x) = L \text{ and } \lim_{x \to c^+} f(x) = L}$

Scenario	Left = Right?	Two-Sided Limit
Both sides agree ( $L_1 = L_2$ )	✓	Exists, equals $L$
Sides disagree ( $L_1 \neq L_2$ )	✗	DNE
One side is $\pm\infty$	✗	DNE (as a finite limit)

AP Tip: On the AP exam, if a problem asks "does the limit exist?", always check both one-sided limits — even if one side seems obvious.

📖 Piecewise Functions & Breakpoints

At each breakpoint (where the rule changes), check both sides:

Example 1: Limit Exists

$g(x) = \begin{cases} x + 3 & x < 2 \\ x^2 & x \geq 2 \end{cases}$

$\lim_{x \to 2^-} g(x) = 2 + 3 = 5$

Since $5 \neq 4$ : $\lim_{x \to 2} g(x)$ does not exist.

Example 2: Limit Exists but Function Disagrees

$f(x) = \begin{cases} x^2 & x < 1 \\ 5 & x = 1 \\ 2x - 1 & x > 1 \end{cases}$

$\lim_{x \to 1^-} f(x) = 1$
$\lim_{x \to 1^+} f(x) = 1$

$\lim_{x \to 1} f(x) = 1$ , but $f(1) = 5 \neq 1$ . The limit exists but the function is at .

🔑 Key Fact: The limit only cares about what happens near the point, not at the point.

Check Your Understanding 🎯

📌 Vertical Asymptotes & One-Sided Behavior

At a vertical asymptote, one-sided limits tell you the direction:

Example: $f(x) = \frac{1}{x - 3}$

Side	Values of $x$	Sign of $x-3$	Limit
$x \to 3^+$	$3.1, 3.01, \ldots$	Positive & small	$+\infty$

Since one side goes to $+\infty$ and the other to $-\infty$ , the two-sided limit DNE.

Example: $g(x) = \frac{1}{(x-3)^2}$

Both sides: $(x-3)^2 > 0$ regardless, so:

$\lim_{x \to 3^+} \frac{1}{(x-3)^2} = +\infty \quad \text{and} \quad \lim_{x \to 3^-} \frac{1}{(x-3)^2} = +\infty$

We write $\lim_{x \to 3} \frac{1}{(x-3)^2} = +\infty$ (both sides agree on going to ).

AP Tip: Even though both sides go to $+\infty$ , this is NOT a finite limit. The limit "does not exist" as a real number. However, writing " $= \infty$ " communicates useful information.

Absolute Value Functions

Recall: $|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$

Example: $\lim_{x \to 0} \frac{|x|}{x}$

From the right ( $x > 0$ ): $\frac{|x|}{x} = \frac{x}{x} = 1$

Since $1 \neq -1$ , the two-sided limit does not exist.

Example: $\lim_{x \to 2} \frac{|x-2|}{x-2}$

From the right ( $x > 2$ ): $|x-2| = x-2$ , so $\frac{x-2}{x-2} = 1$

DNE — same pattern as $\frac{|x|}{x}$ , just shifted.

🔑 Key Fact: $\frac{|\text{expression}|}{\text{expression}}$ always produces $\pm 1$ limits from each side, and the two-sided limit will always be DNE.

Evaluate the One-Sided Limits 🔍

Let $f(x) = \begin{cases} 3x-1 & x < 2 \\ x^2+1 & x \geq 2 \end{cases}$

Compute the One-Sided Limit ✍️

🔑 Key Concept: The Squeeze Theorem (also called the Sandwich or Pinching Theorem) lets you evaluate limits of functions that are trapped between two other functions — even when algebraic techniques fail.

📖 Statement of the Squeeze Theorem

If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $c$ (except possibly at $c$ ), and:

$\boxed{\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L \quad \Rightarrow \quad \lim_{x \to c} f(x) = L}$

The Three Requirements

Requirement	What to Check
1. Lower bound	Find $g(x)$ with $g(x) \leq f(x)$ near $c$
2. Upper bound	Find with near

Intuition: If $f$ is "squeezed" between two functions that both approach $L$ , then $f$ has no escape — it must also approach $L$ .

Common Bounding Facts

Inequality	Use When
$-1 \leq \sin(\text{anything}) \leq 1$	Oscillating $\sin$ factor
$-1 \leq \cos(\text{anything}) \leq 1$

AP Tip: The Squeeze Theorem is the only tool for handling limits with oscillating terms like $\sin(1/x)$ or $\cos(1/x^2)$ .

Classic Oscillation Examples

Example 1: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$

$\sin(1/x)$ oscillates wildly between $-1$ and $1$ as $x \to 0$ . But $x \to 0$ :

$-|x| \leq x\sin\left(\frac{1}{x}\right) \leq |x|$

Since $\lim_{x \to 0} (-|x|) = 0$ and $\lim_{x \to 0} |x| = 0$ :

$\boxed{\lim_{x \to 0} x\sin\left(\frac{1}{x}\right) = 0}$

Example 2: $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)$

$-x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2$

Both $\pm x^2 \to 0$ , so $\lim_{x \to 0} x^2 \cos(1/x) = 0$ .

Example 3: $\lim_{x \to \infty} \frac{\cos x}{x}$

$-\frac{1}{x} \leq \frac{\cos x}{x} \leq \frac{1}{x}$

Both $\pm \frac{1}{x} \to 0$ as $x \to \infty$ , so $\lim_{x \to \infty} \frac{}{}$ .

🔑 Pattern: Oscillating function × vanishing function → limit is 0 (use Squeeze Theorem).

Check Your Understanding 🎯

📌 Proving $\lim_{x \to 0} \frac{\sin x}{x} = 1$

This fundamental result is proved using the Squeeze Theorem and unit circle geometry.

For $0 < x < \frac{\pi}{2}$ , comparing areas of triangles and sectors on the unit circle:

$\boxed{\cos x \leq \frac{\sin x}{x} \leq 1}$

Since $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} 1 = 1$ :

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

Why This Matters

Result	How It's Used
$\lim_{x \to 0} \frac{\sin x}{x} = 1$

AP Tip: You don't need to prove this on the AP exam, but understanding why it's true deepens your grasp of the limit → derivative connection.

Apply the Squeeze Theorem 🔍

Determine each limit.

Apply the Squeeze Theorem ✍️

f is continuous at x = c ⟺ ⎩ ⎨ ⎧ 1. f (c) is defined 2. lim_{x \to c} f (x) exists 3. lim_{x \to c} f (x) = f (c)

If any condition fails → $f$ is discontinuous at $c$ .

Checking Continuity: Systematic Approach

Example: Is $f(x) = \frac{x^2-1}{x-1}$ continuous at $x = 1$ ?

$f(1) = \frac{0}{0}$ — undefined ❌ (Condition 1 fails)

$f$ is discontinuous at $x = 1$ , even though $\lim_{x \to 1} f(x) = 2$ exists.

Example: $g(x) = \begin{cases} x^2 & x \neq 3 \\ 5 & x = 3 \end{cases}$

$g(3) = 5$ ✓
$\lim_{x \to 3} g(x) = 9$ ✓
$9 \neq 5$ ❌ (Condition 3 fails)

AP Tip: On free-response questions, always check all three conditions explicitly. Even if the answer seems obvious, showing the systematic check earns full credit.

Types of Discontinuities

Type	Description	Which Condition Fails?	Example
Removable (hole)	Limit exists but $f(c)$ is missing or wrong	Condition 1 or 3	$\frac{x^2-4}{x-2}$ at $x=2$
Jump	One-sided limits exist but differ	Condition 2	Floor function $\lfloor x \rfloor$ at integers
Infinite	Function → $\pm\infty$	Condition 2	$\frac{1}{x}$ at $x=0$
Oscillating	Function oscillates without settling	Condition 2	$\sin(1/x)$ at $x=0$

Why "Removable" Matters

A removable discontinuity can be "fixed" by redefining $f(c)$ to equal the limit:

$f(x) = \frac{x^2-4}{x-2} \quad \text{has a hole at } x = 2$

Define $f(2) = 4$ (the limit value) → now $f$ is continuous at $x = 2$ .

🔑 Key Fact: A discontinuity is removable if and only if $\lim_{x \to c} f(x)$ exists as a finite number.

Check Your Understanding 🎯

📌 Functions That Are Always Continuous

These functions are continuous on their entire domain:

Function Type	Domain	Continuous On
Polynomials	$(-\infty, \infty)$	All reals
$e^x$ , $a^x$	$(-\infty, \infty)$	All reals
$\sin x$ , $\cos x$	$(-\infty, \infty)$	All reals
$\ln x$	$(0, \infty)$	All positive reals
$\sqrt{x}$	$[0, \infty)$	All non-negative reals
$\frac{1}{x}$	$x \neq 0$	Everywhere except $x =$

Building Continuous Functions

If $f$ and $g$ are continuous at $c$ , then these are also continuous at $c$ :

$f + g$ , $f - g$ , $f \cdot g$
$\frac{f}{g}$ (provided )

🔑 Key Fact: Most functions you encounter are continuous. Discontinuities typically occur at division by zero, piecewise breakpoints, or domain boundaries.

The Intermediate Value Theorem (IVT)

$\boxed{\text{If } f \text{ is continuous on } [a,b], \text{ then } f \text{ takes every value between } f(a) \text{ and } f(b).}$

More precisely: if $N$ is between $f(a)$ and $f(b)$ , then there exists $c \in (a,b)$ with .

Using IVT to Prove a Root Exists

Claim: $x^3 + x - 1 = 0$ has a solution in $[0,1]$ .

Proof:

Let $f(x) = x^3 + x - 1$ (polynomial → continuous on $[0,1]$ ) ✓
and ✓

AP Exam IVT Justification Template

"Since $f$ is continuous on $[a,b]$ and $f(a) = \text{[value]}$ and $f(b) = \text{[value]}$ , and is between and , by the IVT there exists with ."

AP Tip: You MUST state that $f$ is continuous — IVT requires it! Forgetting this is one of the most common point-losing mistakes.

IVT Practice 🎯

Classify the Discontinuities 🔍

Apply the IVT ✍️

🔑 Key Principle: Every limit problem fits one of these patterns. Your job is pattern recognition — the technique follows automatically.

📖 Worked Example 1: Rationalization

$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

Step 1: Direct sub gives $\frac{\sqrt{4}-2}{4-4} = \frac{0}{0}$ → indeterminate

Step 2: Radical in numerator → rationalize (conjugate trick):

$\frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}$

Step 3: Now substitute: $\frac{1}{\sqrt{4}+2} = \frac{1}{4}$

$\boxed{\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \frac{1}{4}}$

Worked Example 2: Trig Limit Manipulation

$\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$

Strategy: Introduce the "missing" denominators to create $\frac{\sin u}{u}$ forms:

$\frac{\sin(3x)}{\sin(5x)} = \frac{\sin(3x)}{3x} \cdot \frac{5x}{\sin(5x)} \cdot \frac{3x}{5x}$

As $x \to 0$ : each $\frac{\sin u}{u} \to 1$ , so the answer is simply the ratio of coefficients:

$\boxed{\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)} = \frac{3}{5}}$

AP Shortcut: $\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b}$ — always the ratio of coefficients.

Practice: Rationalization & Trig 🎯

📖 Worked Example 3: Limits at $-\infty$ with Radicals

$\lim_{x \to -\infty} \frac{2x + 1}{\sqrt{x^2 + 3}}$

The trap: For $x < 0$ , $\sqrt{x^2} = |x| = -x$ , not !

Step 1: Factor $x$ from the numerator and $\sqrt{x^2}$ from the denominator:

$\frac{2x+1}{\sqrt{x^2+3}} = \frac{x(2 + 1/x)}{|x|\sqrt{1 + 3/x^2}}$

Step 2: Since $x < 0$ , we have $|x| = -x$ :

$= \frac{x(2 + 1/x)}{-x\sqrt{1 + 3/x^2}} = \frac{-(2+1/x)}{\sqrt{1+3/x^2}}$

Step 3: As $x \to -\infty$ : $1/x \to 0$ and $3/x^2 \to 0$ :

$\boxed{\lim_{x \to -\infty} \frac{2x+1}{\sqrt{x^2+3}} = \frac{-2}{1} = -2}$

AP Tip: The sign of $\sqrt{x^2}$ is the #1 source of errors on limits at $-\infty$ with radicals. Always ask: "Is positive or negative here?"

Practice: Limits with Radicals 🎯

📌 Complete AP Exam Limit Toolkit

Problem Type	Key Move	Example
$\frac{\sqrt{\ldots} - k}{\text{something}}$	Conjugate multiplication	$\frac{\sqrt{x}-2}{x-4}$
$\frac{\sin(ax)}{bx}$ or $\frac{\sin(ax)}{\sin(bx)}$
Polynomial $\frac{0}{0}$	Factor and cancel	$\frac{x^2-1}{x-1}$
$x \to \pm\infty$ rational	Divide by highest power	$\frac{3x^2+1}{x^2-5} \to 3$
$x \to \pm\infty$ with $\sqrt{\,}$	Use $\sqrt{x^2} =	x
Piecewise or $	\cdot	$
Oscillation	Squeeze Theorem	$x^2\sin(1/x) \to 0$

🔑 Key Fact: On the AP exam, about 3–5 questions test limits directly, plus limits appear implicitly in derivative and integral questions.

Quick Evaluation Drill 🔍

Compute the Limit ✍️

🔑 Key Principle: Mastering limits is the foundation for ALL of calculus — derivatives, integrals, and series all rely on limits.

📖 How Limits Connect to the Rest of AP Calculus

Derivatives Are Limits

$\boxed{f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}}$

Every derivative you compute is secretly a limit! The skills from Parts 1–6 (especially factoring, rationalizing, and trig limits) are essential for computing derivatives from the definition.

Integrals Are Limits

$\boxed{\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x}$

The definite integral is the limit of Riemann sums as the number of rectangles approaches infinity.

L'Hôpital's Rule (Preview)

Later in the course, you'll learn a shortcut for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms:

$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \quad \text{(if conditions are met)}$

For now, the algebraic techniques from this unit are the foundation.

AP Tip: The AP exam tests limits in multiple-choice (computation), free-response (justification with IVT/continuity), and implicitly through derivative and integral problems. Expect 3–5 direct limit questions plus many indirect ones.

Comprehensive Review 🎯

📌 Free-Response Practice: IVT Justification

Problem (AP Style): Let $f$ be the function defined by $f(x) = x^3 - 4x + 2$ .

(a) Show that $f$ has at least one zero in the interval $[-3, 0]$ .

Model Solution:

$f$ is a polynomial, so $f$ is continuous on $[-3, 0]$ .

$f(-3) = (-3)^3 - 4(-3) + 2 = -27 + 12 + 2 = -13$

Grading Rubric (How AP Readers Score This)

Point	Requirement
1	States $f$ is continuous (with reason)
1	Computes $f(a)$ and $f(b)$ correctly
1	Notes $0$ is between and , invokes IVT, states conclusion

AP Tip: Forgetting to state " $f$ is continuous" costs you a point every time. It's the most common mistake on IVT problems.

AP Exam Practice 🎯

Final Review: Name That Technique 🔍

Compute the Limit ✍️

One More Challenge ✍️

\frac{5 x}{2 x} \cdot \frac{s i n ( 5 x ) / ( 5 x )}{s i n ( 2 x ) / ( 2 x )}

⎩⎨⎧0bman±∞if n<mif n=mif n>m

\lim_{x \to 1^+} f(x) = 2(1)-1 = 1

(use the "

x \geq 1

" rule)

lim_{x \to 3^{+}} g (x) = 5

lim_{x \to 1^{+}} h (x) = 4

\lim_{x \to 2^+} g(x) = 2^2 = 4

lim_{x \to 1^{+}} f (x) = 1

+∞andlimx→3−(x−3)21=

From the left (

x < 0

\frac{|x|}{x} = \frac{-x}{x} = -1

From the left (

x < 2

|x-2| = -(x-2)

, so

\frac{-(x-2)}{x-2} = -1

lim_{x \to \infty} \frac{c o s x}{x} = 0

f(g(x))

(composition) — continuous at

c

g

is continuous at

c

and

f

is continuous at

g(c)

Since

f

is continuous on

[0,1]

and

0

is between

f(0) = -1

and

f(1) = 1

, by the IVT there exists

c \in (0,1)

with

f(c) = 0

. ✓

N = \text{[target]}

∣x∣1+3/x2x(2+1/x)

limx→cg′(x)f′(x)(if conditions are met)

$f(0) = 0 - 0 + 2 = 2$

Since $f(-3) = -13 < 0 < 2 = f(0)$ , and $f$ is continuous on $[-3,0]$ , by the Intermediate Value Theorem, there exists $c \in (-3, 0)$ such that $f(c) = 0$ .

Limits & Continuity (AP Calculus AB Unit 1)

Limits & Continuity (AP Calculus AB Unit 1) - Complete Interactive Lesson

Part 1: The Foundation of Calculus

∫ Understanding Limits

Topics in This Part

📖 What Is a Limit?

📖 Evaluating Limits by Direct Substitution

📌 The Indeterminate Form 00\frac{0}{0}00​

When Limits Do Not Exist (DNE)

Key Techniques Summary

Part 2: Mastering Limit Computation

∫ Evaluating Limits Algebraically

Topics in This Part

📖 Special Trig Limits

Part 3: Left-Hand and Right-Hand Limits

∫ One-Sided Limits

Topics in This Part

📖 Left-Hand and Right-Hand Limits

Part 4: Bounding Limits

∫ The Squeeze Theorem

Topics in This Part

Part 5: When Functions Behave Nicely

∫ Continuity & the Intermediate Value Theorem

Topics in This Part

📖 The Three Conditions for Continuity at x=cx = cx=c

Part 6: AP-Level Practice

∫ Problem-Solving Workshop

Strategy Decision Tree

Part 7: Putting It All Together

∫ Review & AP Exam Applications

Complete Limits & Continuity Toolkit

Key Distinction

Graphical Intuition

Functions Where Direct Substitution Always Works

Technique 1: Factoring

Technique 2: Rationalizing (Conjugate Multiplication)

Technique 3: Expanding

nonzero0\frac{\text{nonzero}}{0}0nonzero​ — Not Indeterminate!

Extending the Pattern

General Rule

📖 Limits at Infinity for Rational Functions

Memory Aid

Worked Examples

📌 Limits Involving eee

Important Variants

Piecewise Function Limits & One-Sided Limits

When a Piecewise Limit Fails

The Existence Theorem

📖 Piecewise Functions & Breakpoints

Example 1: Limit Exists

Example 2: Limit Exists but Function Disagrees

📌 Vertical Asymptotes & One-Sided Behavior

Example: f(x)=1x−3f(x) = \frac{1}{x - 3}f(x)=x−31​

Example: g(x)=1(x−3)2g(x) = \frac{1}{(x-3)^2}g(x)=(x−3)21​

Absolute Value Functions

Example: lim⁡x→0∣x∣x\lim_{x \to 0} \frac{|x|}{x}limx→0​x∣x∣​

Example: lim⁡x→2∣x−2∣x−2\lim_{x \to 2} \frac{|x-2|}{x-2}limx→2​x−2∣x−2∣​

📖 Statement of the Squeeze Theorem

The Three Requirements

Common Bounding Facts

Classic Oscillation Examples

Example 1: lim⁡x→0xsin⁡(1x)\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)limx→0​xsin(x1​)

Example 2: lim⁡x→0x2cos⁡(1x)\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)limx→0​x2cos(x

Example 3: lim⁡x→∞cos⁡xx\lim_{x \to \infty} \frac{\cos x}{x}limx→∞​xcosx​

📌 Proving lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1limx→0​xsinx​=1

Why This Matters

Checking Continuity: Systematic Approach

Types of Discontinuities

Why "Removable" Matters

📌 Functions That Are Always Continuous

Building Continuous Functions

The Intermediate Value Theorem (IVT)

Using IVT to Prove a Root Exists

AP Exam IVT Justification Template

📖 Worked Example 1: Rationalization

Worked Example 2: Trig Limit Manipulation

📖 Worked Example 3: Limits at −∞-\infty−∞ with Radicals

📌 Complete AP Exam Limit Toolkit

📖 How Limits Connect to the Rest of AP Calculus

Derivatives Are Limits

📌 The Indeterminate Form $\frac{0}{0}$

📖 The Three Conditions for Continuity at $x = c$

$\frac{\text{nonzero}}{0}$ — Not Indeterminate!

📌 Limits Involving $e$

Example: $f(x) = \frac{1}{x - 3}$

Example: $g(x) = \frac{1}{(x-3)^2}$

Example: $\lim_{x \to 0} \frac{|x|}{x}$

Example: $\lim_{x \to 2} \frac{|x-2|}{x-2}$

Example 1: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$

Example 2: $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)$

Example 3: $\lim_{x \to \infty} \frac{\cos x}{x}$

📌 Proving $\lim_{x \to 0} \frac{\sin x}{x} = 1$

📖 Worked Example 3: Limits at $-\infty$ with Radicals