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.