# UNIT 1: LIMITS AND CONTINUITY

# 1. INTRODUCTION

Calculus isn't just math; it's the "main character" of change. If Algebra is a screenshot, Calculus is the 4K 60fps video. It all starts with the Limit—the ultimate "what if?"

Imagine zooming into a pixel on your screen. You might never actually "touch" the center coordinate, but you can see exactly where the colors are heading as you get closer. In math, a limit asks: "As $x$ gets infinitely close to $c$, what value is $f(x)$ trying to be?" This chapter is your foundation. Master the limit, and you've unlocked the rest of AP Calc. Let's get it.

# 2. ALL KEY CONCEPTS, TERMS, FOUNDATIONAL KNOWLEDGE, AND PRINCIPLES

# 2.1 The Concept of a Limit

• Limit: The value that a function $f(x)$ approaches as the input $x$ approaches some value $c$. It is denoted as $\lim_{x \to c} f(x) = L$.
• One-Sided Limit: The value a function approaches as $x$ approaches $c$ specifically from the left ($x \to c^-$) or from the right ($x \to c^+$).
• Existence of a Limit: A limit $\lim_{x \to c} f(x)$ exists if and only if both one-sided limits exist and are equal: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.

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# 2.2 Continuity

• Continuity at a Point: A function $f(x)$ is continuous at $x = c$ if three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Discontinuity: A "break" in the graph. Types include:
  • Removable (Point) Discontinuity: A "hole" in the graph where the limit exists but the function is undefined or defined elsewhere.
  • Jump Discontinuity: The left and right limits both exist but are not equal.
  • Infinite Discontinuity: The function approaches $\pm \infty$ as $x$ approaches $c$ (Vertical Asymptote).

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# 2.3 Limits Involving Infinity

• Vertical Asymptote: Occurs at $x = c$ if $\lim_{x \to c^+} f(x) = \pm \infty$ or $\lim_{x \to c^-} f(x) = \pm \infty$.
• Horizontal Asymptote: Describes the end-behavior of a function. If $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, then $y = L$ is a horizontal asymptote.

# 2.4 Important Theorems

• The Squeeze Theorem (Sandwich Theorem): If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $c$ (except possibly at $c$) and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$.
• Intermediate Value Theorem (IVT): If $f$ is continuous on the closed interval $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f(c) = k$.

# 3. IN-DEPTH EXPLANATION

# 3.1 Understanding the Limit Notation (Level 0-20)

The notation $\lim_{x \to c} f(x) = L$ is read as "the limit of $f(x)$ as $x$ approaches $c$ is $L$."
It is crucial to understand that $x$ never actually reaches $c$. We are interested in the trend of the $y$-values as the $x$-values get closer and closer to $c$ from both sides.

In the table above, as $x \to 1$, $f(x) \to 2$. Even though $f(1)$ is undefined ($0/0$), the limit exists and is 2.

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# 3.2 Algebraic Techniques for Evaluating Limits (Level 20-50)

When evaluating $\lim_{x \to c} f(x)$, always try Direct Substitution first. If the result is a real number, you are usually finished. If you get an Indeterminate Form such as $\frac{0}{0}$, you must use algebraic manipulation:

1. Factoring: Cancel common factors.
   $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4$$
2. Rationalizing: Use conjugates for square root expressions.
   $$\lim_{x \to 0} \frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \lim_{x \to 0} \frac{x+1-1}{x(\sqrt{x+1}+1)} = \frac{1}{2}$$
3. Complex Fractions: Find a common denominator to simplify.
4. Special Trig Limits:
   $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$

# 3.3 Continuity and its Formal Definition (Level 50-70)

A function is "continuous" if you can draw it without lifting your pencil. However, the AP exam requires the Three-Part Definition of Continuity. For $f(x)$ to be continuous at $x=c$:

1. Existence of Value: $f(c)$ must be defined.
2. Existence of Limit: $\lim_{x \to c} f(x)$ must exist (Left limit = Right limit).
3. Equality: $\lim_{x \to c} f(x) = f(c)$.

If any of these fail, the function is discontinuous. Piecewise functions are common targets for continuity questions. You must ensure the pieces "meet" at the same $y$-value.

# 3.4 Infinite Limits and End Behavior (Level 70-85)

Limits at infinity ($\lim_{x \to \pm \infty} f(x)$) determine the Horizontal Asymptotes.

• If the degree of the numerator $<$ degree of the denominator, the limit is $0$ ($y=0$ is the HA).
• If the degree of the numerator $=$ degree of the denominator, the limit is the ratio of leading coefficients.
• If the degree of the numerator $>$ degree of the denominator, the limit is $\pm \infty$ (No HA).

For Vertical Asymptotes, look for values where the denominator is zero but the numerator is not. The limit as $x$ approaches these values will be $\infty$ or $-\infty$.

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# 3.5 The Squeeze Theorem and IVT (Level 85-100)

The Squeeze Theorem is used when a function is trapped between two other functions whose limits are known.
Example: To find $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$, we know $-1 \leq \sin(\frac{1}{x}) \leq 1$.
Multiplying by $x^2$: $-x^2 \leq x^2 \sin(\frac{1}{x}) \leq x^2$.
Since $\lim_{x \to 0} -x^2 = 0$ and $\lim_{x \to 0} x^2 = 0$, then $\lim_{x \to 0} x^2 \sin(\frac{1}{x}) = 0$.

The Intermediate Value Theorem (IVT) is an existence theorem. It doesn't tell you where a value is, only that it must exist. It is frequently used to prove a function has a root (zero) on an interval $[a, b]$ if $f(a)$ and $f(b)$ have opposite signs.

# 4. EXAMPLES

# Example 1: Direct Substitution

Evaluate $\lim_{x \to 3} (x^2 - 5x + 2)$.
Solution:
Substitute $x = 3$:
$$(3)^2 - 5(3) + 2 = 9 - 15 + 2 = -4$$
Since the function is a polynomial (continuous everywhere), the limit is simply the function value.

# Example 2: Factoring (Indeterminate Form)

Evaluate $\lim_{x \to -3} \frac{x^2 + x - 6}{x + 3}$.
Solution:
Direct substitution gives $\frac{(-3)^2 + (-3) - 6}{-3 + 3} = \frac{0}{0}$. Factor the numerator:
$$\lim_{x \to -3} \frac{(x+3)(x-2)}{x+3} = \lim_{x \to -3} (x-2) = -3 - 2 = -5$$

# Example 3: Rationalizing the Numerator

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

# Example 4: One-Sided Limits

Let $f(x) = \begin{cases} x+1 & x < 2 \ 5 & x = 2 \ x^2-1 & x > 2 \end{cases}$. Find $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$.
Solution:
From the left: $\lim_{x \to 2^-} (x+1) = 3$.
From the right: $\lim_{x \to 2^+} (x^2-1) = 2^2-1 = 3$.
Since $3 = 3$, $\lim_{x \to 2} f(x) = 3$. Note that $f(2)=5$, so the limit exists but the function is not continuous at $x=2$.

# Example 5: Trig Limit Identity

Evaluate $\lim_{x \to 0} \frac{\sin(5x)}{x}$.
Solution:
We use the identity $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Multiply numerator and denominator by 5:
$$\lim_{x \to 0} 5 \cdot \frac{\sin(5x)}{5x} = 5(1) = 5$$

# Example 6: Vertical Asymptote Behavior

Evaluate $\lim_{x \to 2^+} \frac{1}{x-2}$.
Solution:
As $x$ approaches 2 from the right (e.g., 2.1, 2.01), the denominator is a very small positive number.
$$\frac{1}{\text{small positive}} \to \infty$$
Thus, the limit is $\infty$.

# Example 7: Horizontal Asymptote (Same Degree)

Evaluate $\lim_{x \to \infty} \frac{3x^2 - 4x + 5}{2x^2 + 7}$.
Solution:
Divide every term by the highest power of $x$ in the denominator ($x^2$):
$$\lim_{x \to \infty} \frac{3 - \frac{4}{x} + \frac{5}{x^2}}{2 + \frac{7}{x^2}} = \frac{3 - 0 + 0}{2 + 0} = \frac{3}{2}$$

# Example 8: Horizontal Asymptote (Higher Degree Denominator)

Evaluate $\lim_{x \to \infty} \frac{x+5}{x^2-1}$.
Solution:
Since the denominator grows faster than the numerator:
$$\lim_{x \to \infty} \frac{\frac{1}{x} + \frac{5}{x^2}}{1 - \frac{1}{x^2}} = \frac{0}{1} = 0$$

# Example 9: Limit of a Piecewise Function for Continuity

Find the value of $k$ such that $f(x)$ is continuous at $x=3$:
$f(x) = \begin{cases} kx^2 & x \leq 3 \ 2x+k & x > 3 \end{cases}$
Solution:
For continuity, $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)$.
$$k(3)^2 = 2(3) + k \implies 9k = 6 + k \implies 8k = 6 \implies k = \frac{3}{4}$$

# Example 10: Squeeze Theorem Application

Find $\lim_{x \to 0} x^4 \cos(\frac{2}{x})$.
Solution:
$-1 \leq \cos(\frac{2}{x}) \leq 1$
$-x^4 \leq x^4 \cos(\frac{2}{x}) \leq x^4$
$\lim_{x \to 0} -x^4 = 0$ and $\lim_{x \to 0} x^4 = 0$.
By Squeeze Theorem, $\lim_{x \to 0} x^4 \cos(\frac{2}{x}) = 0$.

# Example 11: Intermediate Value Theorem

Show that $f(x) = x^3 + x - 1$ has a zero on $[0, 1]$.
Solution:

1. $f(x)$ is continuous (it's a polynomial).
2. $f(0) = 0^3 + 0 - 1 = -1$.
3. $f(1) = 1^3 + 1 - 1 = 1$.
   Since $f(0) < 0 < f(1)$, by IVT, there exists $c \in (0, 1)$ such that $f(c) = 0$.

# Example 12: Limit at Infinity with Square Roots

Evaluate $\lim_{x \to \infty} \frac{\sqrt{9x^2+1}}{x+1}$.
Solution:
Divide by $x$ (which is $\sqrt{x^2}$ for $x>0$):
$$\lim_{x \to \infty} \frac{\sqrt{\frac{9x^2}{x^2}+\frac{1}{x^2}}}{\frac{x}{x}+\frac{1}{x}} = \frac{\sqrt{9+0}}{1+0} = 3$$

# Example 13: Limit at Negative Infinity

Evaluate $\lim_{x \to -\infty} \frac{\sqrt{4x^2+1}}{x+1}$.
Solution:
Caution: For $x < 0$, $x = -\sqrt{x^2}$.
$$\lim_{x \to -\infty} \frac{\sqrt{x^2(4 + \frac{1}{x^2})}}{x(1 + \frac{1}{x})} = \lim_{x \to -\infty} \frac{|x|\sqrt{4 + \frac{1}{x^2}}}{x(1 + \frac{1}{x})}$$
Since $x \to -\infty$, $|x| = -x$.
$$\lim_{x \to -\infty} \frac{-x\sqrt{4 + 0}}{x(1 + 0)} = -2$$

# Example 14: Removable Discontinuity

Identify the hole in $f(x) = \frac{x^2-x-6}{x-3}$.
Solution:
Factor: $f(x) = \frac{(x-3)(x+2)}{x-3}$. The factor $(x-3)$ cancels.
The hole is at $x=3$. To find the $y$-coordinate: $\lim_{x \to 3} (x+2) = 5$.
Hole at $(3, 5)$.

# Example 15: Infinite Discontinuity (VA)

Find the vertical asymptotes of $f(x) = \frac{x+2}{x^2-4}$.
Solution:
$f(x) = \frac{x+2}{(x+2)(x-2)} = \frac{1}{x-2}$ (for $x \neq -2$).
$x = -2$ is a removable discontinuity (hole).
$x = 2$ is an infinite discontinuity (vertical asymptote).

# Example 16: Complex Fraction Limit

Evaluate $\lim_{x \to 0} \frac{\frac{1}{x+3} - \frac{1}{3}}{x}$.
Solution:
Find common denominator in numerator:
$$\lim_{x \to 0} \frac{\frac{3 - (x+3)}{3(x+3)}}{x} = \lim_{x \to 0} \frac{-x}{3x(x+3)} = \lim_{x \to 0} \frac{-1}{3(x+3)} = -\frac{1}{9}$$

# Example 17: Limit DNE (Oscillation)

Evaluate $\lim_{x \to 0} \sin(\frac{1}{x})$.
Solution:
As $x \to 0$, the value $\frac{1}{x}$ goes to infinity. The sine function oscillates between $-1$ and $1$ infinitely many times as it approaches 0. It does not settle on a single value.
Limit Does Not Exist (DNE).

# Example 18: Non-Removable Jump Discontinuity

Evaluate continuity for $f(x) = \frac{|x|}{x}$ at $x=0$.
Solution:
$\lim_{x \to 0^+} \frac{x}{x} = 1$.
$\lim_{x \to 0^-} \frac{-x}{x} = -1$.
Since $1 \neq -1$, the limit DNE. This is a jump discontinuity.

# Example 19: Finding Constants for Continuity (Advanced)

Find $a$ and $b$ such that $f(x)$ is continuous everywhere:
$f(x) = \begin{cases} 2 & x \leq -1 \ ax+b & -1 < x < 3 \ -2 & x \geq 3 \end{cases}$
Solution:
At $x = -1$: $2 = a(-1) + b \implies -a + b = 2$.
At $x = 3$: $a(3) + b = -2 \implies 3a + b = -2$.
Subtracting the equations: $4a = -4 \implies a = -1$.
Substitute $a$: $-(-1) + b = 2 \implies 1 + b = 2 \implies b = 1$.

# Example 20: Limit with Trig and Factoring

Evaluate $\lim_{x \to 0} \frac{\tan x}{x}$.
Solution:
$$\lim_{x \to 0} \frac{\sin x}{\cos x \cdot x} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right) \cdot \left(\lim_{x \to 0} \frac{1}{\cos x}\right) = 1 \cdot \frac{1}{1} = 1$$

# Example 21: Limit of Composite Functions

Given $\lim_{x \to 2} f(x) = 5$, find $\lim_{x \to 2} [f(x)]^2 - 3$.
Solution:
Using limit properties:
$$(\lim_{x \to 2} f(x))^2 - \lim_{x \to 2} 3 = 5^2 - 3 = 22$$

# Example 22: IVT and Calculator Usage

Prove $x^5 - x^2 + 2x + 3 = 0$ has at least one root.
Solution:
Let $f(x) = x^5 - x^2 + 2x + 3$. $f(x)$ is continuous.
$f(-2) = -32 - 4 - 4 + 3 = -37$.
$f(0) = 3$.
Since $f(-2) < 0 < f(0)$, by IVT, there is a root between $x=-2$ and $x=0$.

# Example 23: Limit at Infinity with Exponential Functions

Evaluate $\lim_{x \to \infty} \frac{e^x}{e^x + 1}$.
Solution:
Divide by $e^x$:
$$\lim_{x \to \infty} \frac{1}{1 + \frac{1}{e^x}} = \frac{1}{1 + 0} = 1$$

# Example 24: Limit at Negative Infinity with Exponentials

Evaluate $\lim_{x \to -\infty} \frac{e^x}{e^x + 1}$.
Solution:
As $x \to -\infty$, $e^x \to 0$.
$$\frac{0}{0+1} = 0$$

# Example 25: Property of Limits - Sum/Difference

If $\lim_{x \to c} f(x) = 4$ and $\lim_{x \to c} g(x) = -2$, find $\lim_{x \to c} [3f(x) + g(x)]$.
Solution:
$3(4) + (-2) = 12 - 2 = 10$.

# Example 26: Limits of Natural Logarithms

Evaluate $\lim_{x \to 0^+} \ln(x)$.
Solution:
As $x$ approaches 0 from the positive side, the natural log function decreases without bound.
Limit = $-\infty$.

# Example 27: Squeeze Theorem with $x \sin(1/x)$

Evaluate $\lim_{x \to 0} x \sin(1/x)$.
Solution:
$-1 \leq \sin(1/x) \leq 1$
$-|x| \leq x \sin(1/x) \leq |x|$
Since $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$, the limit is 0.

# Example 28: Continuity of Piecewise Trig Functions

Find $a$ so $f(x)$ is continuous: $f(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \ a & x = 0 \end{cases}$
Solution:
$\lim_{x \to 0} \frac{\sin x}{x} = 1$.
For continuity, $f(0)$ must equal the limit.
$a = 1$.

# Example 29: Identifying Types of Discontinuity from a Graph

Given a graph with:

• A hole at $x=1$
• A vertical asymptote at $x=2$
• A jump at $x=3$
  Analysis:
• $x=1$: Removable discontinuity.
• $x=2$: Infinite discontinuity.
• $x=3$: Jump discontinuity.

# Example 30: Limit with $1-\cos x$ Identity

Evaluate $\lim_{x \to 0} \frac{1-\cos x}{x^2}$.
Solution:
Multiply by conjugate $1+\cos x$:
$$\lim_{x \to 0} \frac{1-\cos^2 x}{x^2(1+\cos x)} = \lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)}$$
$$\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1+\cos x} = (1)^2 \cdot \frac{1}{2} = \frac{1}{2}$$

# Example 31: End Behavior of Rational Functions (Case: Top Heavy)

Evaluate $\lim_{x \to \infty} \frac{x^3 - 1}{x^2 + 4}$.
Solution:
$$\lim_{x \to \infty} \frac{x - \frac{1}{x^2}}{1 + \frac{4}{x^2}} = \infty$$
No horizontal asymptote exists.

# Example 32: IVT - Specific Value

Does $f(x) = \sqrt{x+1}$ take the value $1.5$ on the interval $[0, 3]$?
Solution:
$f(x)$ is continuous on $[0, 3]$.
$f(0) = \sqrt{1} = 1$.
$f(3) = \sqrt{4} = 2$.
Since $1 < 1.5 < 2$, by IVT, yes, there is some $c \in (0, 3)$ such that $f(c) = 1.5$.

# Example 33: Limit of Absolute Value

Evaluate $\lim_{x \to 2} \frac{|x-2|}{x-2}$.
Solution:
Left: $\lim_{x \to 2^-} \frac{-(x-2)}{x-2} = -1$.
Right: $\lim_{x \to 2^+} \frac{x-2}{x-2} = 1$.
Limit DNE.

# Example 34: Limit involving $\pi$

Evaluate $\lim_{x \to \pi} \frac{\sin x}{x - \pi}$.
Solution:
Let $u = x - \pi$. As $x \to \pi, u \to 0$.
Then $x = u + \pi$.
$$\lim_{u \to 0} \frac{\sin(u + \pi)}{u} = \lim_{u \to 0} \frac{-\sin u}{u} = -1$$

# Example 35: Continuity and Domain

Where is $f(x) = \frac{1}{\sqrt{x^2-1}}$ continuous?
Solution:
The function is continuous on its domain.
$x^2-1 > 0 \implies (x-1)(x+1) > 0$.
$x \in (-\infty, -1) \cup (1, \infty)$.
The function is continuous on these open intervals.

End of Unit 1.