∫ 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 $x = 0$)

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Key Techniques Summary

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

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Match the Technique 🔍

For each limit, select the best first step.

∫ Evaluating Limits Algebraically

Part 2 of 7 — Mastering Limit Computation

1. Special Trig Limits

Two limits you must memorize for the AP exam:

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

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 = $\pm\infty$
  • $\lim_{x \to \infty} \frac{x^3}{x+1} = \infty$

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, $\lim_{x \to 1} f(x) = 1$.

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Limits at Infinity — Quick Reference

Key trig limits to memorize:

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

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Evaluate Each Limit 🔍

∫ 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 $c$ from the left (values less than $c$).

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.

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One-Sided Limit Decision Tree

1. Is the function piecewise? → Use the piece valid on that side
2. Is there a vertical asymptote? → Check sign of expression on that side
3. 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

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Evaluate the One-Sided 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 $c$), and:

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

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

Use it when:

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

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Squeeze Theorem Checklist

To apply the Squeeze Theorem, you need three things:

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

Common bounding facts:

• $-1 \leq \sin(\text{anything}) \leq 1$
• $-1 \leq \cos(\text{anything}) \leq 1$
• $0 \leq |\sin(\text{anything})| \leq 1$

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Apply the Squeeze Theorem 🔍

∫ 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:

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, $f$ 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 $f(c) = 0$.

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Continuity Checklist (AP Exam Format)

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

1. "$f(c)$ is defined and equals ___"
2. "$\lim_{x \to c} f(x)$ exists and equals ___"
3. "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
• "...by the IVT, there exists $c \in (a,b)$ such that $f(c) = N$." ← conclusion

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Classify the Discontinuities 🔍

∫ 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 $\pm\infty$ (or DNE if signs differ by side)
• $\frac{\pm\infty}{\pm\infty}$: 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.

Worked Example 1

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

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

Worked Example 2

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

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

Worked Example 3

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

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

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

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

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Common AP Exam Limit Tricks

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Quick Evaluation 🔍

∫ 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: $x^2 - a^2 = (x-a)(x+a)$
• 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 $f$ continuous at $x = c$? 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): $\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n$

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IVT Application Template

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

Solution framework:

1. State that $f$ is continuous on $[a,b]$ (and say why — polynomial, composition of continuous functions, etc.)
2. Compute $f(a)$ and $f(b)$
3. Note that $0$ is between $f(a)$ and $f(b)$
4. 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 $f(1) = e - 3 + 1 \approx 0.718 > 0$. Since $f$ is continuous and changes sign, $f(c)=0$ for some $c \in (0,1)$.

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Comprehensive Review 🔍