The Danicarl Guide to AP® Calculus BC

Chapter at a Glance

• Chapter: 1
• Strand: Pure — Change
• Difficulty: Level 4 of 5
• Builds on: Limits and continuity, Function analysis

Try This First

Try this short activity before reading the lesson.

> Take $f(x) = \dfrac{x^2 - 1}{x - 1}$. On paper, evaluate it at $x = 0.9$, $0.99$, $1.01$ and $1.1$. Then try $x = 1$. Write one sentence about what the four numbers are heading toward, and one about what happens at $1$ itself.

The Big Idea

A limit is where a function is heading, not where it is. $\frac{x^2-1}{x-1}$ has no value at $x = 1$, but every nearby value sits next to $2$. So $\lim_{x\to 1} \frac{x^2-1}{x-1} = 2$. The whole of calculus is built on this one separation: the value at a point, and the behavior near it.

Most limits on this topic fall to substitution. When substitution gives $\frac{0}{0}$, the expression is hiding a common factor: factor, or rationalize, and cancel it. When $x \to \infty$, only the fastest-growing terms matter.

Continuity is when the two things agree: the limit exists, $f(a)$ exists, and they are equal. Break any one of the three and you have a discontinuity, with a name for each kind.

Words to Know

• limit: $\lim_{x\to a} f(x) = L$ means $f(x)$ gets as close as you like to $L$ when $x$ is close enough to $a$. What $f(a)$ is, or whether it exists, is irrelevant.
• one-sided limit: $\lim_{x\to a^-}$ approaches from the left, $\lim_{x\to a^+}$ from the right. The two-sided limit exists exactly when both exist and agree.
• removable discontinuity: A hole: the limit exists, but $f(a)$ is missing or wrong. Redefine one point and the function is continuous. A jump has two different one-sided limits; an infinite discontinuity has a vertical asymptote.
• Intermediate Value Theorem: If $f$ is continuous on $[a,b]$, it takes every value between $f(a)$ and $f(b)$. Continuous is the hypothesis; without it, the theorem says nothing.

Formulas and Theorems

• Continuity at a point: $\lim_{x\to a} f(x) = f(a)$
• Note: Three things at once: the limit exists, $f(a)$ exists, and they agree.
• Squeeze theorem: $g(x) \le f(x) \le h(x),\ \lim g = \lim h = L \Rightarrow \lim f = L$
• Limit at infinity of a rational function: $\lim_{x\to\infty}\frac{a x^n + \dots}{b x^n + \dots} = \frac{a}{b}$
• Note: Same degree: the ratio of leading coefficients. Top degree lower: $0$. Higher: no finite limit.

Outcomes and Milestones

• Students can: evaluate limits by substitution, factoring, rationalizing and the squeeze theorem; read one-sided limits and limits at infinity from a graph and from algebra; classify discontinuities as removable, jump or infinite; check the three conditions of continuity at a point; apply the Intermediate Value Theorem to show a root exists.
• Watch for: $\lim_{x\to 2}\frac{x^2-4}{x-2}$ called "undefined" because $f(2)$ is; $\lim_{x\to\infty}\frac{3x^2+1}{x^2-5}$ read from the constants instead of the leading terms; IVT applied to a function that is not continuous on the interval.
• Ready when: $\lim_{x\to 3}\frac{x^2-9}{x-3}=6$ found by factoring with the words "the hole is at $3$, the limit is $6$", and one removable and one jump discontinuity classified from a piecewise rule.

Worked Examples and Your Turn

Study each example, then try the paired problem.

Example 1: Substitute first

Problem: Find $\displaystyle\lim_{x\to 2} \frac{x^2 + 3x}{x + 1}$.

Solution:

• Put $x = 2$ in: $\frac{4 + 6}{3} = \frac{10}{3}$. Why: a rational function is continuous wherever its bottom is not zero, and at a continuous point the limit is the value.
• Check the bottom: $2 + 1 = 3 \ne 0$. So substitution was legal and the answer stands.

Answer: $\dfrac{10}{3}$

> Notice: Always try substitution first. Only $\frac{0}{0}$ needs work. A nonzero number over $0$ is a different story (see the edges).

Your Turn:

• Prompt: Find $\displaystyle\lim_{x\to -1} \frac{x^3 + 2}{x^2 + 4}$.
• Answer: Bottom is $5 \ne 0$, so substitute: $\frac{-1 + 2}{5} = \frac{1}{5}$.

Example 2: The hole

Problem: Find $\displaystyle\lim_{x\to 3} \frac{x^2 - x - 6}{x - 3}$.

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Solution:

• Substitute: $\frac{9 - 3 - 6}{0} = \frac{0}{0}$. Why: this form is not an answer, it is an instruction. A zero on top and bottom means both share a factor of $(x - 3)$.
• Factor the top: $x^2 - x - 6 = (x - 3)(x + 2)$. Cancel with the bottom for every $x \ne 3$: the expression equals $x + 2$.
• Now let $x \to 3$: $3 + 2 = 5$. Why: the limit only cares about $x$ near $3$, and there the two expressions are identical.

Answer: $5$

> Notice: Say it in words: the hole is at $3$, the limit is $5$. The function is undefined at $3$. The limit is not.

> Verification & Check: $x = 3.01$: $\frac{9.0601 - 3.01 - 6}{0.01} = \frac{0.0501}{0.01} = 5.01$. Heading to $5$.

Your Turn:

• Prompt: Find $\displaystyle\lim_{x\to -2} \frac{x^2 + 5x + 6}{x + 2}$.
• Answer: $(x+2)(x+3)$ over $(x+2)$ gives $x + 3 \to 1$.

Example 3: Rationalize

Problem: Find $\displaystyle\lim_{x\to 4} \frac{\sqrt{x} - 2}{x - 4}$.

The chapter continues in the book.