AP Calculus AB & BC · 鼎睿学苑

Unit 1: Limits
& Continuity第 1 单元：极限
与连续性

The foundation of calculus. Master how functions behave near a point, formal limit evaluation, continuity, asymptotes, and the Intermediate Value Theorem. 微积分的基础。掌握函数在某点附近的行为、极限的严格求值、连续性、渐近线与介值定理。

10–12% of AP Exam占 AP 考试 10–12% ~22–23 Class Periods约 22–23 课时 16 Topics16 个专题
Topic 1.1专题 1.1

Introducing Calculus: Can Change Occur at an Instant?微积分导引：变化能否在某一瞬间发生？

Core Idea核心思想 Calculus uses limits to understand and model dynamic change. An average rate of change over an interval can approximate the instantaneous rate of change at a point — and as the interval shrinks to zero, the limit gives the exact instantaneous rate. 微积分用极限（limit）来刻画与建模动态变化。区间上的平均变化率（average rate of change）可以近似某一点处的瞬时变化率（instantaneous rate of change）——当区间长度趋于零时，极限给出精确的瞬时变化率。
Average Rate of Change平均变化率
$$ \text{AROC} = \frac{f(b) - f(a)}{b - a} $$
Instantaneous Rate of Change (Preview)瞬时变化率（预览）
$$ \text{IROC at } x = a = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $$

The average rate of change is the slope of a secant line. As $b \to a$, the secant line approaches the tangent line, and the AROC approaches the IROC.平均变化率是割线（secant line）的斜率。当 $b \to a$ 时，割线趋近于切线（tangent line），平均变化率也就趋近于瞬时变化率。

Exam Tip考试提示 AP free-response questions often provide a table of values and ask you to approximate an instantaneous rate of change. Use the average rate of change over the smallest interval containing that point. Always include units. AP 自由回答题常给出一组数值表，并要求近似某点处的瞬时变化率。使用包含该点的最小区间上的平均变化率作答，并务必带上单位。
Worked Example — Approximating IROC例题——近似瞬时变化率

Given:已知： Temperature $T(t)$ in °F at time $t$ (minutes):在时刻 $t$（分钟）的温度 $T(t)$（单位：°F）：

$t$025812
$T(t)$7275808388

Approximate $T'(5)$ using the symmetric interval $[2, 8]$:用对称区间 $[2, 8]$ 近似 $T'(5)$：

$$\begin{aligned} T'(5) &\approx \frac{T(8) - T(2)}{8 - 2} \\[4pt] &= \frac{83 - 75}{6} \\[4pt] &= \frac{8}{6} \approx 1.333 \text{ °F/min} \end{aligned}$$
The average rate of change of $f(x) = x^2$ on $[1, 4]$ is:函数 $f(x) = x^2$ 在 $[1, 4]$ 上的平均变化率为：
1.1
$3$
$5$
$8$
$15$
Correct! $\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5$.正确！$\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5$。
$\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5$. The AROC is the change in output over the change in input.$\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5$。平均变化率即输出的改变量除以输入的改变量。
Topic 1.2专题 1.2

Defining Limits and Using Limit Notation极限的定义与记号

Formal Definition严格定义 Given a function $f$, the limit of $f(x)$ as $x$ approaches $c$ is a real number $R$ if $f(x)$ can be made arbitrarily close to $R$ by taking $x$ sufficiently close to $c$ (but not equal to $c$). 给定函数（function）$f$，当 $x$ 趋于 $c$ 时 $f(x)$ 的极限（limit）为实数 $R$，是指：只要 $x$ 充分接近 $c$（但不等于 $c$），就能使 $f(x)$ 任意接近 $R$。
Limit Notation极限记号
$$ \lim_{x \to c} f(x) = R $$

Key insight: the limit describes what the function approaches — not what the function equals at that point. The value $f(c)$ may or may not equal the limit, or $f(c)$ may not even exist.关键认识：极限描述函数在该点附近所趋近的值，而不是函数在该点处等于多少。$f(c)$ 不一定等于极限，甚至可能不存在。

One-Sided Limits单侧极限 The left-hand limit $\lim_{x \to c^-} f(x)$ considers values of $x$ approaching $c$ from below. The right-hand limit $\lim_{x \to c^+} f(x)$ considers values from above. The two-sided limit exists only when both one-sided limits exist and are equal. 左极限（left-hand limit）$\lim_{x \to c^-} f(x)$ 考察 $x$ 从小于 $c$ 一侧趋于 $c$ 的情形；右极限（right-hand limit）$\lim_{x \to c^+} f(x)$ 考察 $x$ 从大于 $c$ 一侧趋于 $c$ 的情形。双侧极限存在当且仅当两个单侧极限均存在且相等。
Key Relationship关键关系
$$ \lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L \text{ and } \lim_{x \to c^+} f(x) = L $$
Common Exam Trap考试常见陷阱 Students confuse $f(c)$ with $\lim_{x \to c} f(x)$. The limit depends only on function values near $c$, not at $c$ itself. A function can have a limit at a point where it is undefined. 学生常把 $f(c)$ 与 $\lim_{x \to c} f(x)$ 混为一谈。极限只取决于 $c$ 附近的函数值，而与 $c$ 处本身的取值无关。在函数无定义的点处，极限仍可能存在。
Topic 1.3专题 1.3

Estimating Limit Values from Graphs由图像估计极限值

Reading limits from a graph requires careful attention to what the $y$-values approach, not just where the point is plotted.从图像读取极限时，要仔细观察 $y$ 值趋向何处，而不仅仅是图上画出的点的位置。

Key Skills关键技能 Identify left- and right-hand limits from a graph. Look for open circles (holes), filled circles, jump breaks, and vertical asymptotes. A limit might not exist when: the function is unbounded, the function oscillates, or the left- and right-hand limits differ. 从图像辨认左极限与右极限。注意空心圆（图像分析（analyzing graphs）中的"洞"）、实心圆、跳跃间断与垂直渐近线（vertical asymptote）。当函数无界、函数无限振荡或左右极限不相等时，极限可能不存在。
When Limits Do Not Exist (DNE)极限不存在（DNE）的情形 A limit fails to exist when: (1) the left- and right-hand limits are different values (jump), (2) the function increases or decreases without bound (vertical asymptote), or (3) the function oscillates infinitely near the point, e.g. $\lim_{x \to 0} \sin\!\left(\frac{1}{x}\right)$. 极限不存在的情形：(1) 左极限与右极限不相等（跳跃间断点）；(2) 函数无界地增大或减小（垂直渐近线）；(3) 函数在该点附近无限振荡，例如 $\lim_{x \to 0} \sin\!\left(\frac{1}{x}\right)$。
Interactive Limit Explorer交互式极限演示

Drag the slider to move the hole. Watch how $f(x)$ approaches the limit from both sides — the function value at $x = a$ is undefined, but the limit exists.拖动滑块以移动"洞"的位置。观察 $f(x)$ 如何从两侧趋近于极限——在 $x = a$ 处函数无定义，但极限仍然存在。

a = 1.0
Limit =极限 = 2.0
If $\lim_{x \to 3^-} f(x) = 5$ and $\lim_{x \to 3^+} f(x) = 7$, then $\lim_{x \to 3} f(x)$ is:若 $\lim_{x \to 3^-} f(x) = 5$ 且 $\lim_{x \to 3^+} f(x) = 7$，则 $\lim_{x \to 3} f(x)$ 为：
1.3
$5$
$7$
Does not exist不存在
$6$
Correct! Since the left- and right-hand limits are not equal ($5 \neq 7$), the two-sided limit does not exist.正确！因为左极限与右极限不相等（$5 \neq 7$），所以双侧极限不存在。
The two-sided limit exists only when the left- and right-hand limits are equal. Since $5 \neq 7$, the limit does not exist.双侧极限存在当且仅当左极限与右极限相等。由于 $5 \neq 7$，该极限不存在。
Topic 1.4专题 1.4

Estimating Limit Values from Tables由表格估计极限值

Strategy解题策略 To estimate $\lim_{x \to c} f(x)$ from a table, evaluate $f(x)$ at values approaching $c$ from both sides. If $f(x)$ values converge to the same number from both sides, that number is the estimated limit. 由表格估计 $\lim_{x \to c} f(x)$，需在 $c$ 的两侧取若干趋于 $c$ 的 $x$ 值计算 $f(x)$。若 $f(x)$ 的值从两侧都收敛到同一个数，该数即为估计的极限值。
Worked Example — Estimating from a Table例题——由表格估计极限

Estimate $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$估计 $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

$x$1.91.991.9992.0012.012.1
$f(x)$3.93.993.9994.0014.014.1

From the left: $f(x) \to 4$. From the right: $f(x) \to 4$.从左侧：$f(x) \to 4$。从右侧：$f(x) \to 4$。

$$\therefore \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$
Exam Tip考试提示 When building a table on your calculator, choose values very close to $c$ on both sides (like $c \pm 0.001$). Be careful about rounding — don't declare a limit based on values that are far from $c$. 用计算器列表时，在 $c$ 的两侧都选取与 $c$ 非常接近的值（例如 $c \pm 0.001$）。注意四舍五入——不要根据离 $c$ 较远的值就断定极限值。
Practice: If the table shows $f(1.9) = 7.9$, $f(1.99) = 7.99$, $f(2.01) = 8.01$, $f(2.1) = 8.1$, estimate $\lim_{x \to 2} f(x)$.练习：若表格显示 $f(1.9) = 7.9$，$f(1.99) = 7.99$，$f(2.01) = 8.01$，$f(2.1) = 8.1$，估计 $\lim_{x \to 2} f(x)$。
Topic 1.5专题 1.5

Determining Limits Using Algebraic Properties of Limits利用极限的代数性质求极限

Limit Laws极限法则
$$ \lim_{x \to c}[f(x) \pm g(x)] = L \pm M $$ $$ \lim_{x \to c}[f(x) \cdot g(x)] = L \cdot M \qquad \lim_{x \to c}\frac{f(x)}{g(x)} = \frac{L}{M}, \; M \neq 0 $$ $$ \lim_{x \to c}[kf(x)] = kL \qquad \lim_{x \to c}[f(x)]^n = L^n $$
Key Insight关键洞察 Polynomials and rational functions (where the denominator is nonzero) can be evaluated by direct substitution. Always try direct substitution first — if it gives a real number, you're done. 多项式以及分母非零的有理函数都可直接代入（direct substitution）求极限。优先尝试直接代入——若结果是实数即可结束。
Worked Example — Using Limit Laws例题——使用极限法则

Find $\displaystyle\lim_{x \to 3}\bigl[2x^2 + 5x - 1\bigr]$.求 $\displaystyle\lim_{x \to 3}\bigl[2x^2 + 5x - 1\bigr]$。

This is a polynomial — use direct substitution:该式为多项式——直接代入：

$$\begin{aligned} &= 2(3)^2 + 5(3) - 1 \\ &= 2(9) + 15 - 1 \\ &= 18 + 15 - 1 \\ &= 32 \end{aligned}$$
If $\lim_{x \to 2} f(x) = 3$ and $\lim_{x \to 2} g(x) = -1$, then $\lim_{x \to 2} [f(x) \cdot g(x)]$ is:若 $\lim_{x \to 2} f(x) = 3$ 且 $\lim_{x \to 2} g(x) = -1$，则 $\lim_{x \to 2} [f(x) \cdot g(x)]$ 为：
1.5
$-3$
$3$
$2$
DNE
Correct! By the product law, $\lim [f(x) \cdot g(x)] = 3 \cdot (-1) = -3$.正确！由乘积法则，$\lim [f(x) \cdot g(x)] = 3 \cdot (-1) = -3$。
By the product law, $\lim [f(x) \cdot g(x)] = L \cdot M = 3 \cdot (-1) = -3$.由乘积法则，$\lim [f(x) \cdot g(x)] = L \cdot M = 3 \cdot (-1) = -3$。
Topic 1.6专题 1.6

Determining Limits Using Algebraic Manipulation通过代数变形求极限

When direct substitution gives $\frac{0}{0}$ (an indeterminate form), algebraic manipulation is needed to simplify the expression before re-attempting substitution.当直接代入得到 $\frac{0}{0}$，即不定式（indeterminate form）时，需要先做代数变形化简，再重新代入。

Three Key Techniques三大关键技巧 (1) Factoring — factor and cancel common factors. (2) Conjugate multiplication — rationalize expressions with radicals. (3) Trig identities — rewrite using identities before evaluating. (1) 因式分解——分解并消去公因式；(2) 共轭相乘——对含根号的表达式进行有理化；(3) 三角恒等式——用恒等式改写后再求极限。
Worked Example — Factoring例题——因式分解

Find $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.求 $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$。

Direct substitution gives $\frac{9-9}{3-3} = \frac{0}{0}$ — indeterminate. Factor the numerator:直接代入得到 $\frac{9-9}{3-3} = \frac{0}{0}$，为不定式。对分子做因式分解：

$$\begin{aligned} &= \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} \\[4pt] &= \lim_{x \to 3} (x+3) \qquad\text{(cancel } x-3\text{)} \\[4pt] &= 3 + 3 = 6 \end{aligned}$$
Worked Example — Conjugate例题——共轭相乘

Find $\displaystyle\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$.求 $\displaystyle\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$。

Direct substitution gives $\frac{0}{0}$ — multiply by the conjugate:直接代入得到 $\frac{0}{0}$，乘以共轭式：

$$\begin{aligned} &= \lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{(x+4)-4}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} \\[4pt] &= \frac{1}{\sqrt{4}+2} = \frac{1}{4} \end{aligned}$$
Practice: Evaluate $\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}$.练习：求 $\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}$。
Topic 1.7专题 1.7

Selecting Procedures for Determining Limits选择求极限的方法

Decision Flowchart决策流程图 Step 1: Try direct substitution. If you get a number, you're done.
Step 2: If you get $\frac{0}{0}$, try: factoring → conjugate → trig identities → L'Hôpital's (if learned).
Step 3: If you get $\frac{k}{0}$ (where $k \neq 0$), the limit is $\pm \infty$ or DNE — check one-sided limits.
Step 4: If the form is $\frac{\infty}{\infty}$, divide by the highest power of $x$ in the denominator. 第 1 步：先尝试直接代入。若得到一个数，即可结束。
第 2 步：若得到 $\frac{0}{0}$，依次尝试：因式分解 → 共轭相乘 → 三角恒等式 → 洛必达法则（L'Hôpital，若已学）。
第 3 步：若得到 $\frac{k}{0}$（$k \neq 0$），极限为 $\pm \infty$ 或不存在——需检查单侧极限。
第 4 步：若为 $\frac{\infty}{\infty}$ 型，可同时除以分母中 $x$ 的最高次幂。
Exam Tip考试提示 AP questions on Topic 1.7 specifically test your ability to choose the right technique, not just execute it. Practice classifying limit problems before solving them. Ask yourself: "What form does direct substitution give me?" AP 在 1.7 中重点考察你选择合适方法的能力，而不仅仅是执行。先对极限题做分类再下手。问自己："直接代入得到什么形式？"
Direct substitution in $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$ gives $\frac{0}{0}$. The best next step is:对 $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$ 直接代入得 $\frac{0}{0}$，下一步最合适的是：
1.7
Multiply by the conjugate乘以共轭式
Use a table of values使用数值表
Factor both numerator and denominator对分子和分母同时做因式分解
The limit does not exist极限不存在
Correct! Both $x^2 - 4 = (x-2)(x+2)$ and $x^2 - 3x + 2 = (x-2)(x-1)$ share the factor $(x-2)$. After canceling: $\frac{x+2}{x-1}$, which gives $\frac{4}{1} = 4$ at $x = 2$.正确！$x^2 - 4 = (x-2)(x+2)$ 与 $x^2 - 3x + 2 = (x-2)(x-1)$ 同时含有因式 $(x-2)$。消去后得到 $\frac{x+2}{x-1}$，代入 $x = 2$ 得 $\frac{4}{1} = 4$。
When you get $\frac{0}{0}$, factor first. $\frac{(x-2)(x+2)}{(x-2)(x-1)} = \frac{x+2}{x-1}$. At $x = 2$: $\frac{4}{1} = 4$.出现 $\frac{0}{0}$ 时首选因式分解。$\frac{(x-2)(x+2)}{(x-2)(x-1)} = \frac{x+2}{x-1}$，代入 $x = 2$ 得 $\frac{4}{1} = 4$。
Topic 1.8专题 1.8

Determining Limits Using the Squeeze Theorem利用夹逼定理求极限

The Squeeze Theorem夹逼定理
$$ \text{If } g(x) \leq f(x) \leq h(x) \text{ near } c, \text{ and } \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, \text{ then } \lim_{x \to c} f(x) = L $$
Classic Results from the Squeeze Theorem由夹逼定理得出的经典结论 These two limits are proved using the Squeeze Theorem and appear constantly on the AP exam:

$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$     $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ 下面两个极限可由夹逼定理（squeeze theorem）证明，是 AP 考试常考的重要结果：

$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$     $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$
Must-Know Trig Limits必记三角极限
$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 $$
Worked Example — Squeeze Theorem例题——夹逼定理

Show that $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\tfrac{1}{x}\right) = 0$.证明 $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\tfrac{1}{x}\right) = 0$。

We know $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$. Multiply through by $x^2 \ge 0$:已知对所有 $x \neq 0$，$-1 \le \sin(1/x) \le 1$。整体乘以 $x^2 \ge 0$：

$$-x^2 \;\le\; x^2 \sin\!\left(\tfrac{1}{x}\right) \;\le\; x^2$$

Since $\lim_{x \to 0}(-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, by the Squeeze Theorem:由于 $\lim_{x \to 0}(-x^2) = 0$ 且 $\lim_{x \to 0} x^2 = 0$，由夹逼定理：

$$\lim_{x \to 0} x^2 \sin\!\left(\tfrac{1}{x}\right) = 0$$
Worked Example — Using sin(x)/x例题——使用 sin(x)/x

Find $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{5x}$.求 $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{5x}$。

Rewrite to match the standard form $\frac{\sin u}{u} \to 1$:改写以匹配标准形式 $\frac{\sin u}{u} \to 1$：

$$\begin{aligned} &= \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3x}{5x} \\[4pt] &= \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3}{5} \\[4pt] &= 1 \cdot \frac{3}{5} = \frac{3}{5} \end{aligned}$$
Common Exam Trap考试常见陷阱 $\lim_{x \to 0} \frac{\sin x}{x} = 1$ but $\lim_{x \to 0} \frac{x}{\sin x} = 1$ as well. However, $\lim_{x \to 0} \frac{\sin(3x)}{x} = 3$, not 1. Always match the argument: $\frac{\sin(kx)}{kx} \to 1$, so $\frac{\sin(kx)}{x} \to k$. $\lim_{x \to 0} \frac{\sin x}{x} = 1$，同样 $\lim_{x \to 0} \frac{x}{\sin x} = 1$。但是 $\lim_{x \to 0} \frac{\sin(3x)}{x} = 3$，而不是 1。务必保持自变量匹配：$\frac{\sin(kx)}{kx} \to 1$，所以 $\frac{\sin(kx)}{x} \to k$。
Practice: Evaluate $\lim_{x \to 0} \dfrac{\sin(7x)}{x}$.练习：求 $\lim_{x \to 0} \dfrac{\sin(7x)}{x}$。
Topic 1.9专题 1.9

Connecting Multiple Representations of Limits联系极限的多种表示

Four Representations四种表示方式 A limit can be expressed and estimated graphically (reading a graph), numerically (from a table), analytically (algebraic evaluation), and verbally (in words). You must be fluent in all four and able to translate between them. 极限可以通过图像（读图）、数值（由表格）、解析（代数求值）以及语言（用文字描述）四种方式来表达与估计。这四种方式都要熟练，并能在它们之间互相转换。
Exam Tip考试提示 Free-response questions often present data in one representation and ask you to use another. Practice: given a table, sketch the graph. Given a graph, write the limit in analytical notation. Given a formula, describe the limit behavior in words. 自由回答题常用一种表示给出数据，并要求用另一种作答。多多练习：给表格，画图像；给图像，用解析记号写极限；给公式，用语言描述极限行为。
If $\lim_{x \to 2} f(x) = 5$, which is guaranteed?若 $\lim_{x \to 2} f(x) = 5$，下列哪一项必然成立？
1.9
$f(2) = 5$
$f$ is continuous at $x = 2$$f$ 在 $x = 2$ 处连续
$f$ is defined at $x = 2$$f$ 在 $x = 2$ 处有定义
$f(x)$ approaches 5 as $x$ approaches 2当 $x$ 趋于 2 时，$f(x)$ 趋近于 5
Correct! The limit tells us what $f(x)$ approaches — it says nothing about whether $f(2)$ exists, let alone equals 5.正确！极限只告诉我们 $f(x)$ 所趋近的值——并不说明 $f(2)$ 是否存在，更不必等于 5。
The limit describes what $f(x)$ approaches near $x = 2$. It does not guarantee $f(2)$ exists or that $f$ is continuous.极限刻画的是 $x = 2$ 附近 $f(x)$ 所趋近的值，并不保证 $f(2)$ 存在或 $f$ 在该点连续。
Topic 1.10专题 1.10

Exploring Types of Discontinuities探究不连续点的类型

Three Types of Discontinuities不连续点（discontinuity）的三种类型 Removable (hole): The limit exists, but $f(c)$ is either undefined or $f(c) \neq \lim_{x \to c} f(x)$. Can be "fixed" by redefining $f(c)$.

Jump: Both one-sided limits exist but are not equal: $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$. Common in piecewise functions.

Infinite (vertical asymptote): At least one of the one-sided limits is $\pm\infty$. 可去间断点（洞）：极限存在，但 $f(c)$ 无定义或 $f(c) \neq \lim_{x \to c} f(x)$。可通过重新定义 $f(c)$ 来"修补"。

跳跃间断点：两个单侧极限都存在但不相等：$\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$。常见于分段函数。

无穷间断点（垂直渐近线）：至少一个单侧极限为 $\pm\infty$。
Common Exam Trap考试常见陷阱 Don't confuse the type of discontinuity with whether a limit exists. At a removable discontinuity, the limit does exist. At a jump discontinuity, the two-sided limit does not exist. 不要把不连续点的类型与极限是否存在混为一谈。在可去间断点处，极限确实存在；在跳跃间断点处，双侧极限不存在。
$f(x) = \frac{x^2 - 1}{x - 1}$ has what type of discontinuity at $x = 1$?$f(x) = \frac{x^2 - 1}{x - 1}$ 在 $x = 1$ 处的不连续类型为：
1.10
Jump跳跃
Removable可去
Infinite无穷
No discontinuity不存在不连续
Correct! $\frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x + 1$ for $x \neq 1$. The limit is 2, but $f(1)$ is undefined — a hole.正确！当 $x \neq 1$ 时，$\frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x + 1$。极限为 2，但 $f(1)$ 无定义——为一个"洞"。
Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. The limit as $x \to 1$ is 2, but $f(1)$ is undefined. This is a removable discontinuity (hole).因式分解：当 $x \neq 1$ 时，$\frac{(x-1)(x+1)}{x-1} = x+1$。$x \to 1$ 时极限为 2，但 $f(1)$ 无定义。这是可去间断点（洞）。
Topic 1.11专题 1.11

Defining Continuity at a Point某点处连续性的定义

Three Conditions for Continuity at $x = c$$x = c$ 处连续的三个条件
$$ \text{1. } f(c) \text{ is defined} \qquad \text{2. } \lim_{x \to c} f(x) \text{ exists} \qquad \text{3. } \lim_{x \to c} f(x) = f(c) $$

All three conditions must hold. If any one fails, $f$ is discontinuous at $c$.三个条件必须同时满足。任意一个不成立，$f$ 在 $c$ 处就不连续。

Exam Tip — Showing Your Work考试提示——书写过程 When justifying continuity on the AP exam, always explicitly verify all three conditions. Write: "$f(c) = \ldots$", "The limit equals $\ldots$", and "Since $f(c)$ equals the limit, $f$ is continuous at $c$." If discontinuous, state which condition fails. 在 AP 考试中论证连续性时，务必明确地验证全部三个条件。写出："$f(c) = \ldots$"、"极限等于 $\ldots$"、"由于 $f(c)$ 等于极限，$f$ 在 $c$ 处连续。"若不连续，说明哪一条不满足。
Worked Example — Testing Continuity例题——判断连续性

Is $f$ continuous at $x = 2$?$f$ 在 $x = 2$ 处连续吗？

$$f(x) = \begin{cases} x^2 - 1, & x < 2 \\ 5, & x = 2 \\ 2x + 1, & x > 2 \end{cases}$$

Condition 1: $f(2) = 5$ ✓ (defined)条件 1：$f(2) = 5$ ✓（有定义）

Condition 2: Check the limit:条件 2：检查极限：

$$\begin{aligned} \text{Left: } &\lim_{x \to 2^-}(x^2 - 1) = 4 - 1 = 3 \\[4pt] \text{Right: } &\lim_{x \to 2^+}(2x + 1) = 4 + 1 = 5 \end{aligned}$$

Left $\neq$ Right, so the limit does not exist. Condition 2 fails — $f$ is discontinuous at $x = 2$ (jump discontinuity).左 $\neq$ 右，所以极限不存在。条件 2 不满足——$f$ 在 $x = 2$ 处不连续（跳跃间断点）。

Practice: If $f(x) = \begin{cases} x+1, & x \neq 3 \\ 7, & x = 3 \end{cases}$, what is $\lim_{x \to 3} f(x)$?练习：若 $f(x) = \begin{cases} x+1, & x \neq 3 \\ 7, & x = 3 \end{cases}$，则 $\lim_{x \to 3} f(x)$ 为多少？
Topic 1.12专题 1.12

Confirming Continuity over an Interval确认在区间上的连续性

Continuity on an Interval区间上的连续性（continuity） A function is continuous on an open interval $(a, b)$ if it is continuous at every point in the interval. For a closed interval $[a, b]$, we also require that $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$. 函数在开区间 $(a, b)$ 上连续，是指它在区间内每一点都连续。对于闭区间 $[a, b]$，还要求 $\lim_{x \to a^+} f(x) = f(a)$ 且 $\lim_{x \to b^-} f(x) = f(b)$。
Functions Continuous on Their Domains在定义域上处处连续的函数 Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are all continuous at every point in their respective domains. Use this fact to identify intervals of continuity quickly. 多项式函数、有理函数、幂函数、指数函数、对数函数与三角函数在各自的定义域（domain）内每一点都连续。利用这一事实可以迅速判定连续区间。
Topic 1.13专题 1.13

Removing Discontinuities去除不连续点

Removable Discontinuities可去间断点（removable discontinuity） If $\lim_{x \to c} f(x) = L$ exists but $f(c) \neq L$ (or $f(c)$ is undefined), redefining $f(c) = L$ removes the discontinuity. 若 $\lim_{x \to c} f(x) = L$ 存在，但 $f(c) \neq L$（或 $f(c)$ 无定义），则将 $f(c)$ 重新定义为 $L$ 即可去除该不连续。
Worked Example — Piecewise Continuity例题——分段函数的连续性

Find the value of $k$ that makes $f$ continuous at $x = 2$:求使 $f$ 在 $x = 2$ 处连续的 $k$ 值：

$$f(x) = \begin{cases} kx + 1, & x \le 2 \\ x^2 - 1, & x > 2 \end{cases}$$

For continuity, the left and right limits must equal $f(2)$:连续要求左极限与右极限都等于 $f(2)$：

$$\begin{aligned} \text{Left: } &\lim_{x \to 2^-}(kx+1) = 2k + 1 \\[4pt] \text{Right: } &\lim_{x \to 2^+}(x^2 - 1) = 3 \end{aligned}$$

Set equal and solve:令两者相等并求解：

$$2k + 1 = 3 \;\;\Longrightarrow\;\; 2k = 2 \;\;\Longrightarrow\;\; k = 1$$
Practice: Find $k$ so that $f(x) = \begin{cases} 3x + k, & x \leq 1 \\ 5x - 2, & x > 1 \end{cases}$ is continuous at $x = 1$.练习：求 $k$，使得 $f(x) = \begin{cases} 3x + k, & x \leq 1 \\ 5x - 2, & x > 1 \end{cases}$ 在 $x = 1$ 处连续。
Topic 1.14专题 1.14

Connecting Infinite Limits and Vertical Asymptotes无穷极限与垂直渐近线的联系

Vertical Asymptotes垂直渐近线（vertical asymptote） If $\lim_{x \to c^+} f(x) = \pm\infty$ or $\lim_{x \to c^-} f(x) = \pm\infty$, then $x = c$ is a vertical asymptote. This occurs when the denominator approaches $0$ while the numerator approaches a nonzero value. 若 $\lim_{x \to c^+} f(x) = \pm\infty$ 或 $\lim_{x \to c^-} f(x) = \pm\infty$，则 $x = c$ 是垂直渐近线。当分母趋于 $0$ 而分子趋于非零值时即出现这种情形。
Sign Analysis Technique符号分析法 To determine whether the limit is $+\infty$ or $-\infty$, analyze the sign of the function on each side of $c$. Check: is the numerator positive or negative? Is the denominator approaching $0^+$ or $0^-$? Combine the signs. 要判断极限是 $+\infty$ 还是 $-\infty$，分别在 $c$ 两侧分析函数的符号。先看：分子是正还是负？分母趋于 $0^+$ 还是 $0^-$？再综合两个符号。
$\lim_{x \to 0^+} \dfrac{1}{x}$ equals:$\lim_{x \to 0^+} \dfrac{1}{x}$ 等于：
1.14
$+\infty$
$-\infty$
$0$
DNE
Correct! As $x \to 0^+$, both the numerator (1) and denominator ($x$) are positive. Positive divided by a small positive number gives a large positive number, so the limit is $+\infty$.正确！当 $x \to 0^+$ 时，分子（1）与分母（$x$）都为正。正数除以一个小正数得到一个大的正数，故极限为 $+\infty$。
As $x \to 0^+$, $x$ is small and positive, so $1/x$ is large and positive. The limit is $+\infty$.当 $x \to 0^+$ 时，$x$ 为很小的正数，所以 $1/x$ 为很大的正数。极限为 $+\infty$。
Topic 1.15专题 1.15

Connecting Limits at Infinity and Horizontal Asymptotes无穷处的极限与水平渐近线的联系

Horizontal Asymptote Rules for Rational Functions有理函数的水平渐近线（horizontal asymptote）规则 For $f(x) = \frac{a_n x^n + \cdots}{b_m x^m + \cdots}$:

If $n < m$: HA at $y = 0$. Numerator's degree is smaller — denominator "wins."

If $n = m$: HA at $y = \frac{a_n}{b_m}$ (ratio of leading coefficients).

If $n > m$: No horizontal asymptote. The function grows without bound. 对 $f(x) = \frac{a_n x^n + \cdots}{b_m x^m + \cdots}$：

若 $n < m$：水平渐近线为 $y = 0$。分子次数较小——分母"取胜"。

若 $n = m$：水平渐近线为 $y = \frac{a_n}{b_m}$（首项系数之比）。

若 $n > m$：无水平渐近线，函数无界增大。
Interactive End Behavior Explorer交互式末端行为演示

Explore $f(x) = \frac{ax^n}{x^m + 1}$ — adjust the degree of the numerator and denominator to see how horizontal asymptotes depend on the degree comparison.探索 $f(x) = \frac{ax^n}{x^m + 1}$——调整分子与分母的次数，观察水平渐近线如何依赖于次数比较。

a = 2.0
n = 1
m = 1
HA: y = 2.0 (equal degrees)水平渐近线：y = 2.0（次数相同）
Practice: What is $\lim_{x \to \infty} \dfrac{3x^2 + 1}{5x^2 - 2}$?练习：$\lim_{x \to \infty} \dfrac{3x^2 + 1}{5x^2 - 2}$ 等于多少？
Topic 1.16专题 1.16

Working with the Intermediate Value Theorem (IVT)介值定理（IVT）的应用

Intermediate Value Theorem介值定理
$$ \text{If } f \text{ is continuous on } [a, b] \text{ and } d \text{ is between } f(a) \text{ and } f(b), $$ $$ \text{then there exists at least one } c \in (a, b) \text{ such that } f(c) = d. $$
What IVT Says Intuitively介值定理的直观含义 A continuous function can't "skip" values. If $f$ starts below a line and ends above it (or vice versa), $f$ must cross that line at least once. This is used to guarantee the existence of roots. 连续函数不能"跳过"取值。若 $f$ 在起点处低于某条水平线，而在终点处高于该线（或相反），则 $f$ 至少穿越该线一次。常用于保证根的存在性。
IVT Justification Checklist介值定理论证清单 On the AP exam, you MUST state: (1) $f$ is continuous on $[a, b]$ (say why), (2) $d$ is between $f(a)$ and $f(b)$ (compute both), and (3) therefore by the IVT, there exists a $c$ in $(a, b)$ such that $f(c) = d$. Missing any of these loses points. AP 考试中必须明确写出：(1) $f$ 在 $[a, b]$ 上连续（说明原因）；(2) $d$ 位于 $f(a)$ 与 $f(b)$ 之间（两端点的值都要算出来）；(3) 因此由介值定理，存在 $c \in (a, b)$ 使 $f(c) = d$。缺一即扣分。
Worked Example — IVT to Show a Root Exists例题——用介值定理证明根存在

Show that $f(x) = x^3 - 2x - 5$ has a root on $[2,\,3]$.证明 $f(x) = x^3 - 2x - 5$ 在 $[2,\,3]$ 上有根。

Step 1: $f$ is a polynomial, so it is continuous everywhere, including on $[2, 3]$.第 1 步：$f$ 是多项式，所以处处连续，在 $[2, 3]$ 上也连续。

Step 2: Evaluate the endpoints:第 2 步：计算两个端点的值：

$$\begin{aligned} f(2) &= 8 - 4 - 5 = -1 \quad (\text{negative}) \\[4pt] f(3) &= 27 - 6 - 5 = 16 \quad (\text{positive}) \end{aligned}$$

Step 3: Since $f(2) < 0 < f(3)$ and $f$ is continuous on $[2, 3]$, the IVT guarantees at least one $c \in (2, 3)$ where $f(c) = 0$. ✓第 3 步：由于 $f(2) < 0 < f(3)$ 且 $f$ 在 $[2, 3]$ 上连续，由介值定理至少存在一个 $c \in (2, 3)$ 使 $f(c) = 0$。✓

To apply the IVT to show $f(c) = 0$ for some $c$ in $[1, 5]$, you need:要用介值定理证明存在 $c \in [1, 5]$ 使 $f(c) = 0$，需要：
1.16
$f(1) = 0$ or $f(5) = 0$$f(1) = 0$ 或 $f(5) = 0$
$f$ is differentiable on $[1, 5]$$f$ 在 $[1, 5]$ 上可导
$f$ is continuous on $[1, 5]$ and $f(1)$ and $f(5)$ have opposite signs$f$ 在 $[1, 5]$ 上连续，且 $f(1)$ 与 $f(5)$ 异号
$f$ is a polynomial$f$ 是多项式
Correct! The IVT requires continuity on the closed interval and that 0 lies between $f(1)$ and $f(5)$ — which happens when they have opposite signs.正确！介值定理要求函数在闭区间上连续，且 0 位于 $f(1)$ 与 $f(5)$ 之间——即两者异号。
The IVT requires (1) $f$ continuous on $[1, 5]$, and (2) 0 is between $f(1)$ and $f(5)$, i.e., they have opposite signs.介值定理要求：(1) $f$ 在 $[1, 5]$ 上连续；(2) 0 介于 $f(1)$ 与 $f(5)$ 之间，即两者异号。
Review复习

Exam Strategy & Common Pitfalls备考策略与常见陷阱

M
Memorize熟记
  • Three conditions for continuity连续性的三个条件
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$
  • Limit laws (sum, product, quotient)极限法则（和、积、商）
  • IVT statement & conditions介值定理的陈述与前提
  • Rational function HA rules ($n < m$, $n = m$, $n > m$)有理函数水平渐近线规则（$n < m$、$n = m$、$n > m$）
U
Understand理解
  • Why $\lim_{x \to c} f(x) \neq f(c)$ in general为何一般而言 $\lim_{x \to c} f(x) \neq f(c)$
  • How to classify discontinuities如何对不连续点分类
  • When and why each algebraic technique works每种代数技巧何时、为何奏效
  • Squeeze Theorem logic夹逼定理的逻辑
  • Reading limits from graphs and tables从图像与表格读出极限
  • Sign analysis for infinite limits无穷极限的符号分析

Common Pitfalls常见陷阱

Top Student Errors学生最易犯的错误 1. Assuming $\lim_{x \to c} f(x) = f(c)$ without checking continuity.
2. Writing "limit = $\frac{0}{0}$" — $\frac{0}{0}$ is not a limit value, it's a signal to do more work.
3. Forgetting to check both one-sided limits when claiming a limit exists.
4. On IVT problems, failing to state that $f$ is continuous — the hypothesis is required.
5. Confusing "the limit is $\infty$" with "the limit exists" — an infinite limit means the limit DNE as a real number.
6. Incorrectly simplifying $\frac{\sin(kx)}{x}$ — the answer is $k$, not $1$. 1. 不验证连续性就直接假定 $\lim_{x \to c} f(x) = f(c)$。
2. 写出"极限 = $\frac{0}{0}$"——$\frac{0}{0}$ 不是极限值，它只是要求你继续化简的信号。
3. 断言极限存在时漏掉对左右两侧单侧极限的验证。
4. 介值定理题中遗漏"$f$ 连续"这一前提——该前提是必要的。
5. 把"极限为 $\infty$"与"极限存在"混为一谈——无穷极限意味着该极限作为实数不存在（DNE）。
6. 对 $\frac{\sin(kx)}{x}$ 化简错误——答案是 $k$，而不是 $1$。
Review复习

Flashcards闪卡

0 / 8 flipped0 / 8 已翻
What is $\lim_{x \to 0} \frac{\sin x}{x}$?$\lim_{x \to 0} \frac{\sin x}{x} = ?$
$1$
Three conditions for continuity at $x = c$?在 $x = c$ 处连续的三个条件？
1. $f(c)$ defined
2. $\lim_{x \to c} f(x)$ exists
3. $\lim_{x \to c} f(x) = f(c)$1. $f(c)$ 有定义
2. $\lim_{x \to c} f(x)$ 存在
3. $\lim_{x \to c} f(x) = f(c)$
What is a removable discontinuity?什么是可去间断点？
$$\lim_{x\to c} f(x) \text{ exists, but } \neq f(c)$$
IVT requires what hypothesis?介值定理需要哪些前提？
$f$ must be continuous on $[a, b]$, and $d$ must be between $f(a)$ and $f(b)$.$f$ 在 $[a, b]$ 上连续，且 $d$ 介于 $f(a)$ 与 $f(b)$ 之间。
HA when degree of numerator = degree of denominator?分子分母次数相等时的水平渐近线？
$y = \frac{\text{leading coeff of num}}{\text{leading coeff of den}}$
$\frac{0}{0}$ means…$\frac{0}{0}$ 意味着……
$$\text{Indeterminate} \Rightarrow \text{factor, conjugate, or identity}$$
What is the Squeeze Theorem?夹逼定理是什么？
If $g(x) \leq f(x) \leq h(x)$ near $c$ and both outer limits equal $L$, then $\lim_{x \to c} f(x) = L$.若在 $c$ 附近 $g(x) \leq f(x) \leq h(x)$，且两侧极限都等于 $L$，则 $\lim_{x \to c} f(x) = L$。
$\lim_{x \to 0} \frac{1 - \cos x}{x} = \;?$
$0$
Assessment测验

Unit 1 — Practice Quiz第 1 单元——练习测验

Test your understanding. Your score updates in real time at the top of the page.检验你的理解。页面顶部的分数会实时更新。

1. $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$ equals:1. $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$ 等于：
Q1
$2$
$4$
$0$
DNE
Correct! Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$. At $x = 2$: $2 + 2 = 4$.正确！因式分解：$\frac{(x-2)(x+2)}{x-2} = x+2$。代入 $x = 2$：$2 + 2 = 4$。
Factor the numerator: $\frac{(x-2)(x+2)}{x-2} = x + 2$. Substituting $x = 2$ gives $4$.对分子因式分解：$\frac{(x-2)(x+2)}{x-2} = x + 2$。代入 $x = 2$ 得 $4$。
2. $\lim_{x \to 0} \dfrac{\sin(5x)}{x}$ equals:2. $\lim_{x \to 0} \dfrac{\sin(5x)}{x}$ 等于：
Q2
$5$
$1$
$0$
$\frac{1}{5}$
Correct! $\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}$. As $x \to 0$, $\frac{\sin(5x)}{5x} \to 1$, so the answer is $5$.正确！$\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}$。当 $x \to 0$，$\frac{\sin(5x)}{5x} \to 1$，所以答案为 $5$。
Rewrite: $\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x} \to 5 \cdot 1 = 5$.改写：$\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x} \to 5 \cdot 1 = 5$。
3. If $f(x) = \frac{1}{(x - 3)^2}$, then $\lim_{x \to 3} f(x)$ is:3. 若 $f(x) = \frac{1}{(x - 3)^2}$，则 $\lim_{x \to 3} f(x)$ 为：
Q3
$0$
$-\infty$
$\infty$
$1$
Correct! As $x \to 3$, $(x-3)^2 \to 0^+$, so $\frac{1}{(x-3)^2} \to +\infty$ from both sides.正确！当 $x \to 3$，$(x-3)^2 \to 0^+$，所以 $\frac{1}{(x-3)^2}$ 从两侧都趋于 $+\infty$。
Since $(x-3)^2$ is always positive and approaches 0, $\frac{1}{(x-3)^2}$ grows without bound toward $+\infty$.由于 $(x-3)^2$ 始终为正且趋于 0，$\frac{1}{(x-3)^2}$ 无界地增大到 $+\infty$。
4. $\lim_{x \to \infty} \dfrac{2x^3 + x}{4x^3 - 7}$ equals:4. $\lim_{x \to \infty} \dfrac{2x^3 + x}{4x^3 - 7}$ 等于：
Q4
$\infty$
$0$
$2$
$\frac{1}{2}$
Correct! Same degree ($n = m = 3$), so the limit equals the ratio of leading coefficients: $\frac{2}{4} = \frac{1}{2}$.正确！分子分母次数相同（$n = m = 3$），极限等于首项系数之比：$\frac{2}{4} = \frac{1}{2}$。
When the degrees are equal, take the ratio of leading coefficients: $\frac{2}{4} = \frac{1}{2}$.次数相同时，取首项系数之比：$\frac{2}{4} = \frac{1}{2}$。
5. Which condition is NOT required for $f$ to be continuous at $x = c$?5. 下列哪一项不是 $f$ 在 $x = c$ 处连续所需的条件？
Q5
$f(c)$ is defined$f(c)$ 有定义
$f'(c)$ exists$f'(c)$ 存在
$\lim_{x \to c} f(x)$ exists$\lim_{x \to c} f(x)$ 存在
$\lim_{x \to c} f(x) = f(c)$
Correct! Differentiability ($f'(c)$ exists) implies continuity, but continuity does NOT require differentiability. Example: $f(x) = |x|$ is continuous but not differentiable at $x = 0$.正确！可导（$f'(c)$ 存在）蕴含连续，但连续并不要求可导。例如 $f(x) = |x|$ 在 $x = 0$ 处连续但不可导。
The three conditions for continuity are: $f(c)$ defined, limit exists, and limit equals $f(c)$. Differentiability is not required.连续性的三个条件是：$f(c)$ 有定义、极限存在、极限等于 $f(c)$。并不要求可导。
6. $\lim_{x \to 0} \dfrac{1 - \cos(4x)}{x}$ equals:6. $\lim_{x \to 0} \dfrac{1 - \cos(4x)}{x}$ 等于：
Q6
$1$
$0$
$4$
$-4$
Correct! Rewrite as $4 \cdot \frac{1 - \cos(4x)}{4x}$. Since $\frac{1 - \cos u}{u} \to 0$ as $u \to 0$, this equals $4 \cdot 0 = 0$.正确！改写为 $4 \cdot \frac{1 - \cos(4x)}{4x}$。由于当 $u \to 0$ 时 $\frac{1 - \cos u}{u} \to 0$，结果为 $4 \cdot 0 = 0$。
Rewrite as $4 \cdot \frac{1 - \cos(4x)}{4x}$. Since $\lim_{u \to 0} \frac{1 - \cos u}{u} = 0$, the answer is $4 \cdot 0 = 0$.改写为 $4 \cdot \frac{1 - \cos(4x)}{4x}$。因为 $\lim_{u \to 0} \frac{1 - \cos u}{u} = 0$，所以答案为 $4 \cdot 0 = 0$。
7. $f(x) = \frac{x+1}{x^2 - 1}$ has a vertical asymptote at:7. $f(x) = \frac{x+1}{x^2 - 1}$ 的垂直渐近线位于：
Q7
$x = 1$ and $x = -1$$x = 1$ 与 $x = -1$
$x = -1$ only仅 $x = -1$
$x = 0$
$x = 1$ only仅 $x = 1$
Correct! Factor: $\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$ for $x \neq -1$. At $x = -1$ there's a removable discontinuity (hole), and at $x = 1$ there's a vertical asymptote.正确！因式分解：当 $x \neq -1$ 时 $\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$。在 $x = -1$ 处是可去间断点（洞），在 $x = 1$ 处是垂直渐近线。
Factor: $\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$. The $(x+1)$ cancels, leaving a hole at $x = -1$. Only $x = 1$ produces a vertical asymptote.因式分解：$\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$。$(x+1)$ 被约去，$x = -1$ 处只留下一个洞。只有 $x = 1$ 是垂直渐近线。
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Readiness Checklist应考清单

Click each item you've mastered. Aim for 100% before exam day.点击每一项已掌握的条目。考试前争取达到 100%。

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Exam-level practice for this unit — multiple-choice plus extended-response items modeled on the AP rubric. Built for top-score prep; go here after you've worked through the notes and the in-page quizzes above.本单元考试级别的练习——按 AP 评分标准设计的多选题与拓展回答题。专为冲击高分而设；请在完成上方笔记与页内测验后再到此练习。

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AP Calculus AB — Unit 1: Limits and Continuity · 2025 EditionAP 微积分 AB——第 1 单元：极限与连续性 · 2025 版