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Limits & Continuity

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Limits & Continuity - Complete Interactive Lesson

Part 1: The Foundation of Calculus

∫ Understanding Limits

Part 1 of 7 — The Foundation of Calculus

1. What Is a Limit?

A limit describes the value a function approaches as $x$ gets closer and closer to a particular value $c$ . We write:

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

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

Crucial insight: The limit is about where the function is heading, not where it actually is. The function doesn't need to be defined at $x = c$ for the limit to exist.

2. Evaluating Limits by Direct Substitution

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

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

Example: $\lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7$

This works for polynomials, exponentials, and other continuous functions.

3. The Indeterminate Form $\frac{0}{0}$

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

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

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

Multiply by conjugate: $\frac{(x+4)-4}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2} \to \frac{1}{4}$

4. When Limits Do Not Exist

A limit DNE when:

Left-hand and right-hand limits differ
The function grows without bound
The function oscillates (e.g., $\sin(1/x)$ near $x = 0$ )

Check Your Understanding 🎯

Key Techniques Summary

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

Check Your Understanding 🎯

Match the Technique 🔍

For each limit, select the best first step.

Part 2: Mastering Limit Computation

∫ Evaluating Limits Algebraically

Part 2 of 7 — Mastering Limit Computation

1. Special Trig Limits

Two limits you must memorize for the AP exam:

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

Part 3: Left-Hand and Right-Hand Limits

∫ One-Sided Limits

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

1. Definition

The left-hand limit $\lim_{x \to c^-} f(x)$ considers only values of $x$ approaching from the left (values less than ).

Part 4: Bounding Limits

∫ The Squeeze Theorem

Part 4 of 7 — Bounding Limits

1. Statement of the Squeeze Theorem

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

Part 5: When Functions Behave Nicely

∫ Continuity & the Intermediate Value Theorem

Part 5 of 7 — When Functions Behave Nicely

1. Definition of Continuity

A function $f$ is continuous at $x = c$ if all three conditions hold:

$f(c)$ is defined
$\lim_{x \to c} f (x$ exists

Part 6: AP-Level Practice

∫ Problem-Solving Workshop

Part 6 of 7 — AP-Level Practice

Strategy Guide for Limit Problems

Step 1: Try direct substitution. If it works, you're done.

Step 2: Identify the form.

$\frac{0}{0}$ : Try factoring, rationalizing, or trig identities
$\frac{\text{nonzero}}{0}$ : The limit is (or DNE if signs differ by side)

Part 7: Putting It All Together

∫ Review & AP Exam Applications

Part 7 of 7 — Putting It All Together

Complete Limits Toolkit

1. Direct Substitution — Always try first. Works for continuous functions.

2. Algebraic Manipulation — For $\frac{0}{0}$ :

Factor polynomials: $x^2 - a^2 = (x-a)(x+a)$

limx→2x−2(x−2)(x+2)=

AP Tip: $\frac{0}{0}$ does NOT mean DNE. It means "do more algebra."

These appear constantly in disguised forms. For example:

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

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

2. Limits at Infinity

For rational functions as $x \to \infty$ , compare the degrees of numerator and denominator:

Same degree: Limit = ratio of leading coefficients
- $\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - 2} = \frac{3}{5}$
Numerator degree < Denominator degree: Limit = 0
- $\lim_{x \to \infty} \frac{2x}{x^2 + 1} = 0$
Numerator degree > Denominator degree: Limit = $\pm\infty$ $\pm \infty$
- $\lim_{x \to \infty} \frac{x^3}{x+1} = \infty$

3. Limits Involving $e$

The number $e$ is defined by: $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \approx 2.718$

Useful variant: $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

4. Piecewise Function Limits

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

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

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

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

Check Your Understanding 🎯

Limits at Infinity — Quick Reference

Degree Comparison	Result	Memory Aid
deg(top) < deg(bottom)	$0$	"Bottom wins"
deg(top) = deg(bottom)	$\frac{\text{leading coeff top}}{\text{leading coeff bottom}}$	"Tie goes to coefficients"
deg(top) > deg(bottom)	$\pm\infty$	"Top wins"

Key trig limits to memorize:

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

Check Your Understanding 🎯

Evaluate Each Limit 🔍

The right-hand limit $\lim_{x \to c^+} f(x)$ considers only values approaching from the right (values greater than $c$ ).

The two-sided limit exists if and only if both one-sided limits exist and are equal:

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

2. Piecewise Functions

One-sided limits are essential for piecewise functions:

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

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

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

3. Vertical Asymptotes

At a vertical asymptote, one-sided limits reveal the behavior:

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

$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$ (positive values approach from right)
$\lim_{x \to 3^-} \frac{1}{x-3} = -\infty$ (negative values approach from left)

4. Absolute Value Functions

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

$\lim_{x \to 0} \frac{|x|}{x} = \begin{cases} \lim_{x \to 0^+} \frac{x}{x} = 1 \\ \lim_{x \to 0^-} \frac{-x}{x} = -1 \end{cases}$

The two sides disagree, so $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Check Your Understanding 🎯

One-Sided Limit Decision Tree

Is the function piecewise? → Use the piece valid on that side
Is there a vertical asymptote? → Check sign of expression on that side
Is there an absolute value? → Rewrite without $| \cdot |$ using the sign of the expression

Key fact for the AP exam: The two-sided limit exists only when left = right. If the problem asks "does the limit exist," always check both sides.

Common Mistakes

Confusing "the limit is $\infty$ " with "the limit exists" — saying the limit is $\infty$ means it does NOT exist as a finite limit
Forgetting to check both sides at breakpoints of piecewise functions

Check Your Understanding 🎯

Evaluate the One-Sided Limits 🔍

$\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$

then $\lim_{x \to c} f(x) = L$ as well.

Intuition: If $f$ is "squeezed" between two functions that both approach $L$ , then $f$ must also approach $L$ .

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

We know $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$ .

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

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

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

3. When to Use the Squeeze Theorem

The function involves an oscillating factor (like $\sin$ or $\cos$ ) multiplied by something going to 0
You can bound the function between two simpler functions
Direct algebraic techniques don't work

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

Using geometry of the unit circle, one can show that for $0 < x < \frac{\pi}{2}$ :

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

Since $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} 1 = 1$ , the Squeeze Theorem gives us $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

Check Your Understanding 🎯

Squeeze Theorem Checklist

To apply the Squeeze Theorem, you need three things:

✅ A lower bound function $g(x) \leq f(x)$
✅ An upper bound function $f(x) \leq h(x)$
✅ Both bounds approach the same limit: $\lim g = \lim h = L$

Common bounding facts:

$-1 \leq \sin(\text{anything}) \leq 1$
$-1 \leq \cos(\text{anything}) \leq 1$

Check Your Understanding 🎯

Apply the Squeeze Theorem 🔍

lim_{x \to c} f (x)

\lim_{x \to c} f(x) = f(c)

If any condition fails, $f$ has a discontinuity at $c$ .

2. Types of Discontinuities

Removable (hole): The limit exists but $f(c)$ is missing or wrong.

Example: $f(x) = \frac{x^2-1}{x-1}$ at $x=1$ . The limit is 2, but $f(1)$ is undefined.

Jump: The one-sided limits exist but are not equal.

Example: $f(x) = \begin{cases} 1 & x < 0 \\ 2 & x \geq 0 \end{cases}$

Infinite (vertical asymptote): The function approaches $\pm\infty$ .

Example: $f(x) = \frac{1}{x}$ at $x = 0$ .

3. Continuity on an Interval

$f$ is continuous on $[a,b]$ if:

$f$ is continuous at every point in $(a,b)$
$\lim_{x \to a^+} f(x) = f(a)$ (right-continuous at left endpoint)
$\lim_{x \to b^-} f(x) = f(b)$ (left-continuous at right endpoint)

Key fact: Polynomials, $e^x$ , $\sin x$ , $\cos x$ , and $\ln x$ (on its domain) are continuous everywhere they are defined.

4. The Intermediate Value Theorem (IVT)

If $f$ is continuous on $[a,b]$ and $N$ is any value between $f(a)$ and $f(b)$ , then there exists at least one $c \in (a,b)$ such that $f(c) = N$ .

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

$f(0) = -1 < 0$
$f(1) = 1 > 0$
Since $f$ is continuous and changes sign, by IVT there exists $c \in (0,1)$ with $f(c) = 0$ .

Check Your Understanding 🎯

Continuity Checklist (AP Exam Format)

To show $f$ is continuous at $x=c$ , state:

" $f(c)$ is defined and equals ___"
" $\lim_{x \to c} f(x)$ exists and equals ___"
"Since $\lim_{x \to c} f(x) = f(c)$ , $f$ is continuous at $c$ "

IVT on the AP Exam

Standard IVT justification:

"Since $f$ is continuous on $[a,b]$ ..." ← must state continuity
"...and $f(a) = \text{value}$ and $f(b) = \text{value}$ ..." ← state the function values

Check Your Understanding 🎯

Classify the Discontinuities 🔍

\frac{\pm\infty}{\pm\infty}

: Divide top and bottom by highest power of

x

Step 3: For piecewise or absolute value, check both one-sided limits.

Step 4: For oscillating factors, try the Squeeze Theorem.

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

Direct sub: $\frac{0}{0}$ . Rationalize:

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

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

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

Rewrite: $\frac{\sin(3x)}{\sin(5x)} = \frac{\sin(3x)}{3x} \cdot \frac{5x}{\sin(5x)} \cdot \frac{3x}{5x} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}$

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

For $x < 0$ : $\sqrt{x^2} = |x| = -x$ . Factor:

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

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

Check Your Understanding 🎯

Common AP Exam Limit Tricks

Problem Type	Key Move
$\frac{\sqrt{\text{stuff}} - \text{number}}{\text{something}}$	Multiply by conjugate
$\frac{\sin(ax)}{\sin(bx)}$	Rewrite using $\frac{\sin u}{u} \to 1$
$\frac{f(x)}{g(x)}$ as $x \to \pm\infty$	Divide by highest power of
$\frac{0}{0}$ with polynomials	Factor!
Function with $	x

Check Your Understanding 🎯

Quick Evaluation 🔍

Rationalize radicals: multiply by the conjugate

Simplify complex fractions

3. Special Trig Limits:

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

4. Limits at Infinity — Compare degrees of top and bottom.

5. Squeeze Theorem — For oscillating functions bounded by converging functions.

6. One-Sided Limits — Check both sides for piecewise, absolute value, or asymptotes.

AP Exam Format Notes

Multiple choice: Often tests recognition of technique + computation
Free response: May ask you to justify continuity or apply IVT with complete sentences
Common FRQ prompt: "Is $f$ $f$ continuous at $x = c$ $x = c$ ? Justify your answer."
- You must check all three conditions explicitly

Connections to Coming Topics

Limits lead directly to:

Derivatives (Part 2): $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Integrals (Part 4): $\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum f(x_i)\Delta x$
Series (BC only): $\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n$

Check Your Understanding 🎯

IVT Application Template

Problem: Show that $f(x) = 0$ has a solution on $[a,b]$ .

Solution framework:

State that $f$ is continuous on $[a,b]$ (and say why — polynomial, composition of continuous functions, etc.)
Compute $f(a)$ and $f(b)$
Note that $0$ is between $f(a)$ and $f(b)$
Conclude by IVT: there exists $c \in (a,b)$ with $f(c) = 0$

Example: $e^x = 3 - x$ on $[0, 1]$

Let $f(x) = e^x - 3 + x$ . Then $f(0) = 1-3+0 = -2 < 0$ and . Since is continuous and changes sign, for some .

Check Your Understanding 🎯

Comprehensive Review 🔍

\lim_{x \to 0} \frac{1-\cos x}{x} = 0

\lim_{x \to 0} \frac{\tan x}{x} = 1

0 \leq ∣ sin (anything) ∣ \leq 1

"...by the IVT, there exists

c \in (a,b)

such that

f(c) = N

." ← conclusion

f(1) = e - 3 + 1 \approx 0.718 > 0

Limits & Continuity

2. Limits at Infinity

3. Limits Involving $e$

4. Piecewise Function Limits

Limits at Infinity — Quick Reference

2. Piecewise Functions

3. Vertical Asymptotes

4. Absolute Value Functions

One-Sided Limit Decision Tree

Common Mistakes

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

3. When to Use the Squeeze Theorem

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

Squeeze Theorem Checklist

2. Types of Discontinuities

3. Continuity on an Interval

4. The Intermediate Value Theorem (IVT)

Continuity Checklist (AP Exam Format)

IVT on the AP Exam

Worked Example 1

Worked Example 2

Worked Example 3

Common AP Exam Limit Tricks

AP Exam Format Notes

Connections to Coming Topics

IVT Application Template

Limits & Continuity

Limits & Continuity - Complete Interactive Lesson

Part 1: The Foundation of Calculus

∫ Understanding Limits

1. What Is a Limit?

2. Evaluating Limits by Direct Substitution

3. The Indeterminate Form 00\frac{0}{0}00​

4. When Limits Do Not Exist

Key Techniques Summary

Part 2: Mastering Limit Computation

∫ Evaluating Limits Algebraically

1. Special Trig Limits

Part 3: Left-Hand and Right-Hand Limits

∫ One-Sided Limits

1. Definition

Part 4: Bounding Limits

∫ The Squeeze Theorem

1. Statement of the Squeeze Theorem

Part 5: When Functions Behave Nicely

∫ Continuity & the Intermediate Value Theorem

1. Definition of Continuity

Part 6: AP-Level Practice

∫ Problem-Solving Workshop

Strategy Guide for Limit Problems

Part 7: Putting It All Together

∫ Review & AP Exam Applications

Complete Limits Toolkit

2. Limits at Infinity

3. Limits Involving eee

4. Piecewise Function Limits

Limits at Infinity — Quick Reference

2. Piecewise Functions

3. Vertical Asymptotes

4. Absolute Value Functions

One-Sided Limit Decision Tree

Common Mistakes

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

3. When to Use the Squeeze Theorem

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

Squeeze Theorem Checklist

2. Types of Discontinuities

3. Continuity on an Interval

4. The Intermediate Value Theorem (IVT)

Continuity Checklist (AP Exam Format)

IVT on the AP Exam

Worked Example 1

Worked Example 2

Worked Example 3

Common AP Exam Limit Tricks

AP Exam Format Notes

Connections to Coming Topics

IVT Application Template

3. The Indeterminate Form $\frac{0}{0}$

3. Limits Involving $e$

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

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