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Introduction to Limits and Calculus极限与微积分入门

Practice Questions · AP Calc AB-Feeder · ON MCV4U · BC Calculus 12 · AB Math 31练习题集 · AP Calc AB 衔接 · 安大略 MCV4U · 卑诗 Calculus 12 · 阿尔伯塔 Math 31

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Math 31-style阿尔伯塔 Math 31 风格 Honors荣誉级
Name:姓名：Date:日期：
PART I  ·  SHORT RESPONSE第一部分  ·  短答题AP-feeder MCQ + ON/BC short answer · 18 marksAP 衔接选择题 + 安/卑省考短答 · 共 18 分

Section A · Short ResponseA 部分 · 短答题

Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show enough work in the margin that a marker could verify. For limits, justify with a one-line reason (table, factor-and-cancel, direct substitution, or asymptote dominance). No calculator on Q1–Q4; calculator permitted on Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写出足以让阅卷人核对的过程。求极限时用一行说明所采用的依据（数表、因式约分、直接代入或渐近线主导）。Q1–Q4 不可使用计算器；Q5 可用计算器。

Q1EASY易Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §1 Intuitive Limits直观的极限 · AP Calc AB Unit 1 [3 marks][3 分]

The table below gives selected values of a function $f$ near $x = 2$:下表给出函数 $f$ 在 $x = 2$ 附近的若干取值：

$x$:   $1.9$   $1.99$   $1.999$    $2.001$   $2.01$   $2.1$
$f(x)$: $4.41$   $4.9401$   $4.994001$    $5.006001$   $5.0601$   $5.61$

Which value best estimates $\displaystyle\lim_{x\to 2} f(x)$?下列哪一项最佳地估计了 $\displaystyle\lim_{x\to 2} f(x)$？

  1. (A) $4$
  2. (B) $5$
  3. (C) $5.006$
  4. (D) The limit does not exist.极限不存在。
Q2EASY易Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Algebraic Limits代数法求极限 · AP Calc AB Unit 1 [3 marks][3 分]

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

  1. (A) $0$
  2. (B) $3$
  3. (C) $6$
  4. (D) The limit does not exist (the function is undefined at $x = 3$).极限不存在（因为函数在 $x = 3$ 处无定义）。
Q3MEDIUM中Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Power Rule幂法则 · AP Calc AB Unit 2 [3 marks][3 分]

If $f(x) = 4 x^{5} - 3 x^{2} + 7 x - 11$, then $f'(x) = $设 $f(x) = 4 x^{5} - 3 x^{2} + 7 x - 11$，则 $f'(x) = $

  1. (A) $20 x^{4} - 6 x + 7$
  2. (B) $20 x^{5} - 6 x^{2} + 7$
  3. (C) $4 x^{4} - 3 x + 7$
  4. (D) $20 x^{4} - 6 x + 7 - 11$
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 One-Sided Limits & Continuity单侧极限与连续性 · MCV4U Strand A [4 marks][4 分]

Consider the piecewise function考虑分段函数 $$f(x) \;=\; \begin{cases} 2x + 1, & x < 1 \\ 5, & x = 1 \\ x^{2} + 2, & x > 1 \end{cases}$$

(a) Compute the left-hand limit $\displaystyle\lim_{x \to 1^{-}} f(x)$ and the right-hand limit $\displaystyle\lim_{x \to 1^{+}} f(x)$.求左极限 $\displaystyle\lim_{x \to 1^{-}} f(x)$ 与右极限 $\displaystyle\lim_{x \to 1^{+}} f(x)$。 [1]
(b) State whether $\displaystyle\lim_{x \to 1} f(x)$ exists, and justify in one line.判断 $\displaystyle\lim_{x \to 1} f(x)$ 是否存在，并用一行说明理由。 [2]
(c) State whether $f$ is continuous at $x = 1$, and identify which of the three continuity conditions fails (if any).判断 $f$ 在 $x = 1$ 处是否连续，并指出三条连续性条件中哪一条不满足（若有）。 [1]
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Limits at Infinity无穷大极限 · BC Calc 12 [5 marks][5 分]

Consider the rational function $g(x) = \dfrac{3 x^{2} - 5 x + 1}{2 x^{2} + 4}$.考虑有理函数 $g(x) = \dfrac{3 x^{2} - 5 x + 1}{2 x^{2} + 4}$。

(a) Compute $\displaystyle\lim_{x \to \infty} g(x)$ by dividing numerator and denominator by the highest power of $x$ in the denominator. Show the dominant-term step.将分子分母同除以分母中 $x$ 的最高次幂，求 $\displaystyle\lim_{x \to \infty} g(x)$。写出主导项步骤。 [2]
(b) Compute $\displaystyle\lim_{x \to -\infty} g(x)$. State whether the answer differs from part (a) and why.求 $\displaystyle\lim_{x \to -\infty} g(x)$。说明结果是否与 (a) 不同，并解释原因。 [2]
(c) State the equation of the horizontal asymptote of $y = g(x)$.写出 $y = g(x)$ 的水平渐近线方程。 [1]
PART II  ·  EXTENDED RESPONSE第二部分  ·  简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Extended ResponseB 部分 · 简答题

Show every algebraic step. State the limit law or technique invoked (direct substitution, factor and cancel, conjugate, dominant-term, definition of derivative, power rule) before substituting. For "from first principles" items, you must write the definition $f'(x) = \lim_{h \to 0} (f(x+h) - f(x))/h$ before reaching for the power rule. No calculator on Q6–Q9.每一步代数运算都要写出。在代入之前先注明所用极限法则或技巧（直接代入、因式约分、共轭、主导项、导数定义、幂法则）。对于"用定义求导"的题，必须先写出 $f'(x) = \lim_{h \to 0} (f(x+h) - f(x))/h$，再用幂法则。Q6–Q9 不可使用计算器。

Q6MEDIUM中Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §4 Derivative from Definition用定义求导数 · AP Calc AB Unit 2 / MCV4U Strand B [8 marks][8 分]

Let $f(x) = x^{2} - 4 x + 5$.设 $f(x) = x^{2} - 4 x + 5$。

(a) Write the limit definition of the derivative $f'(x)$.写出导数 $f'(x)$ 的极限定义。 [1]
(b) Compute $f(x + h)$ in expanded form, then form the difference quotient $\dfrac{f(x + h) - f(x)}{h}$ and simplify so that the factor of $h$ in the numerator cancels cleanly.展开 $f(x + h)$，然后写出差商 $\dfrac{f(x + h) - f(x)}{h}$ 并化简，使分子中的 $h$ 因子能与分母干净地约去。 [3]
(c) Take the limit as $h \to 0$ to obtain $f'(x)$.令 $h \to 0$ 取极限，得到 $f'(x)$。 [2]
(d) Verify your answer by applying the power rule directly to $f(x)$. Then evaluate $f'(3)$ and interpret it as the slope of the tangent line to $y = f(x)$ at the point $(3, f(3))$.用幂法则直接对 $f(x)$ 求导以验证答案。然后求 $f'(3)$，并将其解释为曲线 $y = f(x)$ 在点 $(3, f(3))$ 处的切线斜率。 [2]
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Power Rule & Linearity幂法则与线性性 · MCV4U Strand B [8 marks][8 分]

Differentiate each function with respect to $x$. State the power rule, sum / difference rule, and constant-multiple rule as you use them. No need to simplify beyond combining like terms.对每个函数关于 $x$ 求导。在使用幂法则、求和 / 差法则与常数倍法则时，请逐条注明。只需合并同类项即可，无需进一步化简。

(a) $p(x) = 6 x^{4} - 2 x^{3} + 9 x - 14$. [2]
(b) $q(x) = \dfrac{5}{x^{2}} + 8 \sqrt{x}$. (Rewrite with rational exponents first.)（请先改写为有理数指数形式。） [3]
(c) For $p(x)$ in part (a), find the equation of the tangent line at $x = 1$.对 (a) 中的 $p(x)$，求其在 $x = 1$ 处的切线方程。 [3]
Q8HARD难 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §2 Algebraic Limits代数法求极限 · BC Calc 12 / AB Math 31 [8 marks][8 分]

Evaluate each limit. State the technique (direct substitution, factor and cancel, or conjugate) before computing.求下列各极限。在计算之前先注明所用技巧（直接代入、因式约分或共轭法）。

(a) $\displaystyle\lim_{x \to 4} \frac{x^{2} - 16}{x^{2} - x - 12}$. [3]
(b) $\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$ (use the conjugate).（使用共轭法）。 [3]
(c) $\displaystyle\lim_{x \to 2} \frac{x^{3} - 8}{x - 2}$ (recognize the difference-of-cubes pattern).（识别立方差的因式分解模式）。 [2]
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §3 Continuity at a Parameter参数取值下的连续性 · AP Calc AB Unit 1 / BC Calc 12 [11 marks][11 分]

Consider the piecewise function考虑分段函数 $$h(x) \;=\; \begin{cases} \dfrac{x^{2} - a^{2}}{x - a}, & x \ne a \\ b, & x = a \end{cases}$$ where $a$ and $b$ are real parameters.其中 $a$、$b$ 为实参数。

(a) For $x \ne a$, simplify $\dfrac{x^{2} - a^{2}}{x - a}$ by factoring the numerator and cancelling.当 $x \ne a$ 时，对分子作因式分解并约分，化简 $\dfrac{x^{2} - a^{2}}{x - a}$。 [2]
(b) Compute $\displaystyle\lim_{x \to a} h(x)$ in terms of $a$.求 $\displaystyle\lim_{x \to a} h(x)$ （用 $a$ 表示）。 [2]
(c) State the three conditions for continuity of $h$ at $x = a$. Use them to determine the value of $b$ (in terms of $a$) that makes $h$ continuous at $x = a$.写出 $h$ 在 $x = a$ 处连续的三条条件。借此确定使 $h$ 在 $x = a$ 处连续的 $b$ 值（用 $a$ 表示）。 [3]
(d) Take $a = 5$. With the value of $b$ from part (c), classify the original "hole" at $x = 5$ as either a removable discontinuity (now removed) or an essential discontinuity, and justify in one line.取 $a = 5$。用 (c) 中得到的 $b$ 值，将原本在 $x = 5$ 处的"洞"分类为可去间断点（现已修补）或本性间断点，并用一行说明理由。 [2]
(e) Sketch (in your work area) the graph of $h$ for $a = 5, b = 10$, showing the line $y = x + 5$ with the point $(5, 10)$ filled in. State the slope of the line.在答题区内画出 $a = 5, b = 10$ 时 $h$ 的图像，画出直线 $y = x + 5$，并将点 $(5, 10)$ 标为实心点。写出直线的斜率。 [2]
PART III  ·  MODELING / APPLIED第三部分  ·  建模 / 应用题Universal · 28 marks通用 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Name your variables (with units) before writing equations. Set up your model, solve, and conclude with a one-sentence answer in context. Always include $+ C$ on indefinite integrals; the constant of integration is worth its own A1 on every AP Calc AB free-response item. Calculator permitted throughout Part III.在列方程前先为变量命名并标注单位。建立模型、求解，并用一句话给出情境中的结论。不定积分必须加上 $+ C$；积分常数在 AP Calc AB 简答题中独占一分（A1）。第三部分全程可使用计算器。

Q10MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Tangent Line & Instantaneous Rate切线与瞬时变化率 · MCV4U Strand A & B [9 marks][9 分]

A drone's height above the ground, in metres, after $t$ seconds is modeled by $s(t) = -5 t^{2} + 40 t$ for $0 \le t \le 8$.一架无人机在 $t$ 秒后的离地高度（米）由模型 $s(t) = -5 t^{2} + 40 t$ 给出，其中 $0 \le t \le 8$。

(a) Compute the average rate of change of $s$ on the interval $[1, 3]$, with units.求 $s$ 在区间 $[1, 3]$ 上的平均变化率，并写出单位。 [2]
(b) Compute the instantaneous rate of change $s'(t)$ using the power rule, and evaluate $s'(2)$. Interpret $s'(2)$ in context, with units.用幂法则求瞬时变化率 $s'(t)$，并求出 $s'(2)$。结合情境解释 $s'(2)$ 的含义，并标出单位。 [2]
(c) Find the time at which the drone's instantaneous vertical velocity is zero. Interpret the result in context (peak height).求无人机瞬时垂直速度为零的时刻。结合情境解释结果（即达到最大高度的时刻）。 [2]
(d) Find the equation of the tangent line to $y = s(t)$ at $t = 1$, in slope-intercept form.求曲线 $y = s(t)$ 在 $t = 1$ 处的切线方程（斜截式）。 [3]
Q11MEDIUM中Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 Antiderivative & Indefinite Integral反导数与不定积分 · AP Calc AB Unit 6 [10 marks][10 分]

A particle moves along a straight line with velocity $v(t) = 3 t^{2} - 4 t + 2$ (in metres per second) at time $t$ (in seconds). The particle's position at time $t = 0$ is $s(0) = 5$ metres.一个质点沿直线运动，时刻 $t$（秒）时的速度为 $v(t) = 3 t^{2} - 4 t + 2$（米/秒）。初始位置 $s(0) = 5$ 米。

(a) Find the general antiderivative $s(t)$ of $v(t)$, including the constant of integration $C$. State the power rule applied in reverse: $\displaystyle\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$.求 $v(t)$ 的一般反导数 $s(t)$，含积分常数 $C$。注明所用的逆向幂法则：$\displaystyle\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$（$n \ne -1$）。 [3]
(b) Use the initial condition $s(0) = 5$ to determine $C$, and write the particular antiderivative $s(t)$.利用初值条件 $s(0) = 5$ 确定 $C$，写出特解 $s(t)$。 [2]
(c) Verify by differentiation that $s'(t) = v(t)$ for your particular antiderivative from part (b).通过求导验证 (b) 中的特解满足 $s'(t) = v(t)$。 [2]
(d) Compute the indefinite integral $\displaystyle\int \left( 4 x^{3} - \frac{6}{x^{2}} + 5 \right) dx$. (Rewrite $1/x^{2}$ as $x^{-2}$ before applying the reverse power rule. Do not forget $+ C$.)计算不定积分 $\displaystyle\int \left( 4 x^{3} - \frac{6}{x^{2}} + 5 \right) dx$。（在使用逆向幂法则前先将 $1/x^{2}$ 改写为 $x^{-2}$。别忘了 $+ C$。） [3]
Q12HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §7 Definite Integral as Signed Area定积分即带符号面积 · AP Calc AB Unit 6 / FTC [9 marks][9 分]

Consider the function $f(x) = 4 - x^{2}$ on the interval $[-2, 3]$.考虑函数 $f(x) = 4 - x^{2}$ 在区间 $[-2, 3]$ 上的性质。

(a) Sketch the graph of $y = f(x)$ over $[-2, 3]$ in your work area. Mark the $x$-intercepts and shade the region between the graph and the $x$-axis on this interval.在答题区内画出 $y = f(x)$ 在 $[-2, 3]$ 上的图像。标出 $x$ 轴交点，并对图像与 $x$ 轴之间的区域阴影。 [1]
(b) The definite integral $\displaystyle\int_{a}^{b} f(x) \, dx$ represents the signed area between the graph and the $x$-axis on $[a, b]$ (area above the axis counts positively; area below counts negatively). Split the interval $[-2, 3]$ at the $x$-intercept $x = 2$ and describe in words the sign of the area on $[-2, 2]$ and on $[2, 3]$.定积分 $\displaystyle\int_{a}^{b} f(x) \, dx$ 表示图像与 $x$ 轴在 $[a, b]$ 之间的带符号面积（轴上方为正、下方为负）。在 $x$ 轴交点 $x = 2$ 处将区间 $[-2, 3]$ 拆分，并用文字说明 $[-2, 2]$ 与 $[2, 3]$ 上面积的符号。 [2]
(c) Using the Fundamental Theorem of Calculus, compute $\displaystyle\int_{-2}^{2} (4 - x^{2}) \, dx$ via an antiderivative $F$ of $f$ and the rule $\int_{a}^{b} f \, dx = F(b) - F(a)$.利用微积分基本定理，借助 $f$ 的反导数 $F$ 与公式 $\int_{a}^{b} f \, dx = F(b) - F(a)$，计算 $\displaystyle\int_{-2}^{2} (4 - x^{2}) \, dx$。 [3]
(d) Compute $\displaystyle\int_{2}^{3} (4 - x^{2}) \, dx$ the same way. The result should be negative, confirming the signed-area picture from part (b).用同样方法计算 $\displaystyle\int_{2}^{3} (4 - x^{2}) \, dx$。结果应为负，与 (b) 中带符号面积的描述相符。 [1]
(e) Combine parts (c) and (d) to compute $\displaystyle\int_{-2}^{3} (4 - x^{2}) \, dx$. State whether the result represents "net signed area" or "total geometric area between the curve and the $x$-axis", and explain in one line.合并 (c)、(d) 的结果，计算 $\displaystyle\int_{-2}^{3} (4 - x^{2}) \, dx$。说明该结果代表"净带符号面积"还是"曲线与 $x$ 轴之间的总几何面积"，并用一行解释。 [2]
🇺🇸 US Common Core美国共同核心Not in CCSSM. Closest CCSSM standard: HSF-IF.B.6 (average rate of change). Whole unit flagged Honors as AP Calc AB feeder; AP Calc AB CED Units 1 (Limits & Continuity), 2 (Differentiation), 6 (Integration) are the proper home.不在 CCSSM 内。最接近的标准：HSF-IF.B.6（平均变化率）。整个单元标 荣誉级，作为 AP Calc AB 衔接；其正式归属为 AP Calc AB CED 的 Unit 1（极限与连续）、Unit 2（求导）、Unit 6（积分）。
🇨🇦 Ontario · MCV4U安大略 · MCV4UStrand A Rate of Change (limits, average vs instantaneous rate, secants & tangents). Strand B Derivatives (definition, power rule, sum & constant-multiple rules, derivative graphs). Antiderivatives / integrals not in MCV4U.单元 A 变化率（极限、平均与瞬时变化率、割线与切线）。单元 B 导数（定义、幂法则、求和与常数倍法则、导数图像）。反导数 / 积分不在 MCV4U 内。
🇨🇦 BC Calc 12 · 🇨🇦 AB Math 31卑诗 Calc 12 · 阿尔伯塔 Math 31BC Calculus 12 (optional Grade 12): limits, derivatives, integrals (citation at course level – calc12_extract.md pending). AB Math 31 (optional Grade 12): same scope as BC Calc 12 (citation at course level – math31_extract.md pending).BC Calculus 12（12 年级选修）：极限、导数、积分（按课程级引用 – calc12_extract.md 待补）。AB Math 31（12 年级选修）：范围与 BC Calc 12 一致（按课程级引用 – math31_extract.md 待补）。

Full 6-region Syllabus Map (with row-by-row scope, defer / skip, and source extracts) lives in ../Study Guides/Unit_15_Introduction_to_Limits_and_Calculus.html.完整的六地区大纲对照表（逐行范围、缓 / 跳与原文摘录）见 ../Study Guides/Unit_15_Introduction_to_Limits_and_Calculus.html。