Limits & Continuity (AP Calculus AB Unit 1)

∫ Understanding Limits

Part 1 of 7 — The Foundation of Calculus

Topics in This Part

🔑 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 $c$:

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

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

Key Distinction

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

Graphical Intuition

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.

📖 Evaluating Limits by Direct Substitution

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

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

Functions Where Direct Substitution Always Works

Worked Example:

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

Substitute $x = 2$: $8 - 8 + 7 = 7$ ✓

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

Check Your Understanding 🎯

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

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

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

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.

More Practice 🎯

When Limits Do Not Exist (DNE)

A limit does not exist when:

$\frac{\text{nonzero}}{0}$ — 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.

Key Techniques Summary

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!

Match the Technique 🔍

For each limit, select the best first step.

Compute the Limit ✍️

∫ Evaluating Limits Algebraically

Part 2 of 7 — Mastering Limit Computation

Topics in This Part

🔑 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:

$\boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0}$

Extending the Pattern

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

General Rule

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

Worked Example:

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

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

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

AP Tip: The limit $\frac{e^x - 1}{x} \to 1$ is really $f'(0)$ where $f(x) = e^x$. 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 "$x < 1$" rule)
• $\lim_{x \to 1^+} f(x) = 2(1)-1 = 1$ (use the "$x \geq 1$" 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 $g$ 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 ✍️

∫ One-Sided Limits

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

Topics in This Part

🔑 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 $c$:

$\boxed{\lim_{x \to c^-} f(x) = L_1}$

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

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

The Existence Theorem

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

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

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

• $\lim_{x \to 2^-} g(x) = 2 + 3 = 5$
• $\lim_{x \to 2^+} g(x) = 2^2 = 4$

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 not continuous at $x = 1$.

🔑 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: $f(x) = \frac{1}{x - 3}$

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 $+\infty$).

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 ($x > 0$): $\frac{|x|}{x} = \frac{x}{x} = 1$
• From the left ($x < 0$): $\frac{|x|}{x} = \frac{-x}{x} = -1$

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 $\frac{x-2}{x-2} = 1$
• From the left ($x < 2$): $|x-2| = -(x-2)$, 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 ✍️

∫ The Squeeze Theorem

Part 4 of 7 — Bounding Limits

Topics in This Part

🔑 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

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

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 $\lim_{x \to \infty} \frac{\cos x}{x} = 0$.

🔑 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

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 ✍️

∫ Continuity & the Intermediate Value Theorem

Part 5 of 7 — When Functions Behave Nicely

Topics in This Part

🔑 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 $x = c$

$\boxed{f \text{ is continuous at } x = c \iff \begin{cases} 1. \ f(c) \text{ is defined} \\ 2. \ \lim_{x \to c} f(x) \text{ exists} \\ 3. \ \lim_{x \to c} f(x) = f(c) \end{cases}}$

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 $x = 1$?

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

1. $g(3) = 5$ ✓
2. $\lim_{x \to 3} g(x) = 9$ ✓
3. $9 \neq 5$ ❌ (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

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:

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 $g(c) \neq 0$)
• $f(g(x))$ (composition) — continuous at $c$ if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$

🔑 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 $f(c) = N$.

Using IVT to Prove a Root Exists

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

Proof:

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

AP Exam IVT Justification Template

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

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 ✍️

∫ Problem-Solving Workshop

Part 6 of 7 — AP-Level Practice

Strategy Decision Tree

🔑 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 $x$!

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 $x$ positive or negative here?"

Practice: Limits with Radicals 🎯

📌 Complete AP Exam Limit Toolkit

🔑 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 ✍️

∫ Review & AP Exam Applications

Part 7 of 7 — Putting It All Together

Complete Limits & Continuity Toolkit

🔑 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 $[-3, 0]$.

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

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

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 $f(c) = 0$.

Grading Rubric (How AP Readers Score This)

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 ✍️