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

Companion to the Practice Set · Mark-by-mark walkthroughs · AP Calc AB-feeder / MCV4U / BC Calc 12 / AB Math 31 styles练习题配套答案 · 逐分讲解 · AP Calc AB 衔接 / MCV4U / BC Calc 12 / AB Math 31 风格

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
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解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 · Worked SolutionsA 部分 · 详细解答

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]

Estimate $\lim_{x \to 2} f(x)$ from the symmetric table around $x = 2$.根据 $x = 2$ 附近的对称数表估计 $\lim_{x \to 2} f(x)$。

Answer:答案：  (B)  $5$

(a) Read both sides of the table同时读表的两侧 M1·A1·A1

As $x$ creeps up on $2$ from the left ($1.9 \to 1.99 \to 1.999$), the $f$-values run $4.41 \to 4.9401 \to 4.994001$, climbing toward $5$. As $x$ creeps down on $2$ from the right ($2.001 \to 2.01 \to 2.1$), the $f$-values run $5.006001 \to 5.0601 \to 5.61$, falling toward $5$. Both one-sided trends converge on the same value, so the two-sided limit exists and equals $\mathbf{5}$. (The table is consistent with $f(x) = x^{2} + 1$, where $f(2) = 5$, but the limit argument does not need to know the formula.)当 $x$ 从左侧逼近 $2$（$1.9 \to 1.99 \to 1.999$）时，$f$ 的取值为 $4.41 \to 4.9401 \to 4.994001$，向 $5$ 攀升。当 $x$ 从右侧逼近 $2$（$2.001 \to 2.01 \to 2.1$）时，$f$ 的取值为 $5.006001 \to 5.0601 \to 5.61$，向 $5$ 下降。两侧趋势收敛于同一值，故双侧极限存在且等于 $\mathbf{5}$。（数表与 $f(x) = x^{2} + 1$ 一致，此时 $f(2) = 5$，但求极限的论证不需要知道具体公式。）
Why the wrong choices fail.干扰项分析。
  • (A) $4$, reads only the far-left value $f(1.9) = 4.41$ and rounds aggressively; ignores that the entries get closer to $5$, not $4$, as $x$ gets closer to $2$.：只读了最左端的 $f(1.9) = 4.41$ 并粗暴取整；忽略了随着 $x$ 趋近 $2$，数据是在向 $5$（而非 $4$）靠拢。
  • (C) $5.006$, copies the right-most-of-the-close-in entries verbatim instead of recognizing the trend toward a clean value of $5$. A limit is the value the function approaches, not the closest tabulated entry.：直接照抄最接近 $2$ 的右侧数据，没有看出数据是在向干净的 $5$ 靠拢。极限是函数所趋近的值，不是最接近的表中条目。
  • (D) The limit does not exist极限不存在, would require the two one-sided trends to disagree (or to oscillate). Here both sides agree on $5$, so the limit exists.：这要求两侧趋势不一致（或振荡）。此处两侧都收敛到 $5$，故极限存在。
A symmetric table around $c$ is a yes/no test for the two-sided limit.围绕 $c$ 的对称数表，就是判断双侧极限的是 / 否检验。 If the left-side column and the right-side column close in on the same value, the limit exists and equals that value. If they close in on different values, the limit does not exist (jump discontinuity). If the values blow up or oscillate without settling, the limit does not exist either. The Study Guide $\S 1$ frames this as "$x$ gets arbitrarily close to $c$ from either side; the limit does not care what $f(c)$ is." For AP Calc AB Unit 1, every tabular-limit MCQ rewards the student who reads both columns before answering.若左右两列收敛于同一值，则极限存在且等于该值。若收敛于不同值，则极限不存在（跳跃间断）。若数值发散或振荡不定，极限同样不存在。学习指南 $\S 1$ 概括为"$x$ 从两侧任意逼近 $c$；极限不关心 $f(c)$ 是什么"。在 AP Calc AB Unit 1 中，每一道数表型极限选择题都奖励先看两侧再作答的学生。
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 $\lim_{x \to 3} \dfrac{x^{2} - 9}{x - 3}$.求 $\lim_{x \to 3} \dfrac{x^{2} - 9}{x - 3}$。

Answer:答案：  (C)  $6$

(a) Direct substitution gives $0/0$, an indeterminate form直接代入得 $0/0$，为不定式 M1

Plugging $x = 3$ into $\dfrac{x^{2} - 9}{x - 3}$ gives $\dfrac{9 - 9}{3 - 3} = \dfrac{0}{0}$, which is indeterminate, not "undefined-so-the-limit-does-not-exist". The $0/0$ form is the trigger to factor and cancel.将 $x = 3$ 代入 $\dfrac{x^{2} - 9}{x - 3}$ 得 $\dfrac{9 - 9}{3 - 3} = \dfrac{0}{0}$，这是不定式，并非"无定义 = 极限不存在"。$0/0$ 就是因式分解与约分的信号。

(b) Factor and cancel, then substitute因式分解、约分，再代入 A1·A1

The numerator is a difference of squares: $x^{2} - 9 = (x - 3)(x + 3)$. So for $x \ne 3$,分子是平方差：$x^{2} - 9 = (x - 3)(x + 3)$。故对 $x \ne 3$： $$ \frac{x^{2} - 9}{x - 3} \;=\; \frac{(x - 3)(x + 3)}{x - 3} \;=\; x + 3. $$ Therefore $\lim_{x \to 3} \dfrac{x^{2} - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$. Matches option (C).因此 $\lim_{x \to 3} \dfrac{x^{2} - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$。对应选项 (C)。
Why the wrong choices fail.干扰项分析。
  • (A) $0$, mistakes the indeterminate $0/0$ for "$0$ divided by anything is $0$". The numerator and denominator are both heading to zero; that requires factoring, not arithmetic.：把不定式 $0/0$ 错当成"$0$ 除以任何数都是 $0$"。分子与分母都在趋于 $0$，需要的是因式分解，不是算术。
  • (B) $3$, sloppy substitution into the simplified $x + 3$ at $x = 0$ instead of $x = 3$; or, alternatively, drops the $+ 3$ after cancellation.：把化简后的 $x + 3$ 误代入 $x = 0$ 而非 $x = 3$；或在约分后丢掉了 $+ 3$。
  • (D) The limit does not exist极限不存在, confuses "$f(3)$ undefined" with "$\lim_{x \to 3} f(x)$ undefined". The Study Guide $\S 1$ "hole example" is exactly this point: the limit can exist even when $f(c)$ does not.：把 "$f(3)$ 无定义"与 "$\lim_{x \to 3} f(x)$ 不存在"混为一谈。学习指南 $\S 1$ 的"洞"例子正是为了厘清这点：即便 $f(c)$ 无定义，极限仍可以存在。
$0/0$ is the algebra-do-something signal, not a final answer.$0/0$ 是"动手做代数"的信号，不是终答。 Every algebraic-limits problem starts with direct substitution. If you get a real number, you are done (limit laws plus continuity of polynomials guarantee it). If you get $0/0$, that's the prompt to reach for the next tool: factor and cancel, conjugate, combine fractions, then take the limit. The Study Guide $\S 2$ ranks these three moves in order: factor-cancel is by far the most common on AP Calc AB Unit 1. Other indeterminate forms ($\infty/\infty$, $\infty - \infty$, $0 \cdot \infty$) all have their own algebraic moves; the AB exam reuses these patterns relentlessly.每道代数极限题都先从直接代入开始。若得到实数，工作完成（极限运算法则与多项式连续性保证）。若得 $0/0$，则提示动用下一个工具：因式约分、共轭法、合并分式，再取极限。学习指南 $\S 2$ 给出三项顺序：因式约分在 AP Calc AB Unit 1 中出现频率最高。其余不定式（$\infty/\infty$、$\infty - \infty$、$0 \cdot \infty$）各有专属代数手法；AB 考试会反复使用这些套路。
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]

$f(x) = 4 x^{5} - 3 x^{2} + 7 x - 11$; find $f'(x)$.设 $f(x) = 4 x^{5} - 3 x^{2} + 7 x - 11$，求 $f'(x)$。

Answer:答案：  (A)  $20 x^{4} - 6 x + 7$

(a) Apply power, constant-multiple, sum / difference rules term-by-term逐项使用幂法则、常数倍法则、求和 / 差法则 M1·A1·A1

Use $(x^{n})' = n x^{n - 1}$ on each term, with $(c f)' = c f'$ and $(f + g)' = f' + g'$:对每项使用 $(x^{n})' = n x^{n - 1}$，配合 $(c f)' = c f'$ 与 $(f + g)' = f' + g'$：
  • $(4 x^{5})' = 4 \cdot 5 x^{4} = 20 x^{4}$.
  • $(-3 x^{2})' = -3 \cdot 2 x = -6 x$.
  • $(7 x)' = 7 \cdot 1 \cdot x^{0} = 7$.
  • $(-11)' = 0$ (derivative of a constant is zero).（常数的导数为零）。
Sum: $f'(x) = 20 x^{4} - 6 x + 7$, matching option (A).求和：$f'(x) = 20 x^{4} - 6 x + 7$，对应选项 (A)。
Why the wrong choices fail.干扰项分析。
  • (B) $20 x^{5} - 6 x^{2} + 7$, multiplies the coefficient by the exponent but forgets to subtract $1$ from each exponent. The power rule $n x^{n - 1}$ has two pieces; both must apply.：把系数乘了指数，却忘了把每个指数减 $1$。幂法则 $n x^{n - 1}$ 有两部分，两步都要做。
  • (C) $4 x^{4} - 3 x + 7$, subtracts $1$ from each exponent but forgets the coefficient-times-exponent step. A common slip when sketching the rule from memory.：每个指数减 $1$，却漏了"系数乘指数"那一步。凭记忆默写法则时常见的疏漏。
  • (D) $20 x^{4} - 6 x + 7 - 11$, leaves the constant $-11$ in the answer. The derivative of any constant is $0$, so $-11$ vanishes; keeping it is the single most common power-rule slip on the AP Calc AB MCQ section.：把常数 $-11$ 留在答案里。任意常数的导数都为 $0$，故 $-11$ 消失；保留它是 AP Calc AB 选择题部分最常见的幂法则失误。
Constants vanish; linear terms become their slope; everything else loses one degree.常数消失；线性项变为斜率；其余各项次数降 $1$。 The mental scan for any polynomial: each $a x^{n}$ becomes $a n x^{n - 1}$, the linear term $b x$ becomes $b$, and every additive constant disappears. So the derivative of any degree-$5$ polynomial is a degree-$4$ polynomial with no constant inherited from the original constant. The Study Guide $\S 5$ frames this as "linearity of differentiation": $(c f + d g)' = c f' + d g'$. AP Calc AB Unit 2 builds product, quotient, and chain rules on top of this, but the linearity over sums and constant multiples never goes away.对任意多项式做心算：每个 $a x^{n}$ 变为 $a n x^{n - 1}$，线性项 $b x$ 变为 $b$，所有加法常数消失。故任意五次多项式的导数为四次多项式，且不会继承原常数项。学习指南 $\S 5$ 概括为"求导的线性性"：$(c f + d g)' = c f' + d g'$。AP Calc AB Unit 2 在此之上构建积、商、链式法则，但求和与常数倍的线性性始终不变。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §3 One-Sided Limits & Continuity · MCV4U Strand A [4 marks]

$f(x) = 2x + 1$ for $x < 1$; $f(1) = 5$; $f(x) = x^{2} + 2$ for $x > 1$. (a) One-sided limits. (b) Two-sided limit. (c) Continuity at $x = 1$.

Answer:  (a) $\lim_{x \to 1^{-}} f = 3$, $\lim_{x \to 1^{+}} f = 3$  ·  (b) $\lim_{x \to 1} f = 3$ (both sides agree)  ·  (c) Discontinuous at $x = 1$: condition 3 fails ($\lim \ne f(1)$).

(a) Plug each branch into its own side M1

Left side ($x < 1$ branch): $\lim_{x \to 1^{-}} (2 x + 1) = 2(1) + 1 = 3$.
Right side ($x > 1$ branch): $\lim_{x \to 1^{+}} (x^{2} + 2) = (1)^{2} + 2 = 3$.
Both one-sided limits equal $3$. The $f(1) = 5$ assignment plays no role here; the limit is about how $f$ behaves near $1$, not at $1$.

(b) Two-sided limit exists iff the two one-sided limits agree A1·R1

Since $\lim_{x \to 1^{-}} f(x) = 3 = \lim_{x \to 1^{+}} f(x)$, the two-sided limit exists and equals $3$: $$ \lim_{x \to 1} f(x) \;=\; 3. $$

(c) Continuity at $x = 1$: three conditions A1

A function is continuous at $c$ iff all three of the following hold:
  1. $f(c)$ is defined — here $f(1) = 5$. $\checkmark$
  2. $\lim_{x \to c} f(x)$ exists — here it equals $3$. $\checkmark$
  3. $\lim_{x \to c} f(x) = f(c)$ — here $3 \ne 5$. $\times$ FAILS.
So $f$ is discontinuous at $x = 1$. The discontinuity is removable: redefining $f(1) = 3$ would patch the function and make it continuous.
One-sided limits ignore $f(c)$; the third continuity condition is the only one that asks about it. MCV4U Strand A's classic ON-marker mistake: students compute the one-sided limits correctly and then write "continuous because both sides agree". That conflates two distinct ideas. "Both sides agree" guarantees the limit exists. Continuity additionally requires the limit to equal $f(c)$. The Study Guide $\S 3$ lists the three conditions explicitly; provincial markers split the (c)-mark cleanly into "limit exists" (A1) and "limit equals function value" (A1). Get both wrong and you lose 2 marks; get only the limit-existence right and you lose 1.
Q5MEDIUM 🇨🇦 BC BC Provincial-style §3 Limits at Infinity · BC Calc 12 [5 marks]

$g(x) = \dfrac{3 x^{2} - 5 x + 1}{2 x^{2} + 4}$. (a) $\lim_{x \to \infty}$. (b) $\lim_{x \to -\infty}$. (c) Horizontal asymptote.

Answer:  (a) $\dfrac{3}{2}$  ·  (b) $\dfrac{3}{2}$ (same; even-degree dominant terms)  ·  (c) $y = \dfrac{3}{2}$

(a) Divide top and bottom by $x^{2}$ M1·A1

The highest power of $x$ in the denominator is $x^{2}$. Divide every term in numerator and denominator by $x^{2}$: $$ g(x) \;=\; \frac{3 x^{2} - 5 x + 1}{2 x^{2} + 4} \;=\; \frac{3 - \dfrac{5}{x} + \dfrac{1}{x^{2}}}{2 + \dfrac{4}{x^{2}}}. $$ As $x \to \infty$, the terms $\dfrac{5}{x}, \dfrac{1}{x^{2}}, \dfrac{4}{x^{2}}$ all tend to $0$, leaving the dominant ratio: $$ \lim_{x \to \infty} g(x) \;=\; \frac{3 - 0 + 0}{2 + 0} \;=\; \frac{3}{2}. $$

(b) Same calculation works for $x \to -\infty$ A1·R1

The $\dfrac{5}{x}, \dfrac{1}{x^{2}}, \dfrac{4}{x^{2}}$ terms still vanish (they go to $0$ from a different side, but the limit is still $0$). So $\lim_{x \to -\infty} g(x) = \dfrac{3}{2}$ as well. The answer matches part (a) because the dominant terms of both numerator ($3 x^{2}$) and denominator ($2 x^{2}$) have the same even degree, so the sign of $x$ at infinity does not matter.

(c) Horizontal asymptote A1

Both limits at infinity agree on $\dfrac{3}{2}$, so $y = g(x)$ has a single horizontal asymptote: $$ y \;=\; \frac{3}{2}. $$
For rational functions: degree top vs degree bottom determines the horizontal-asymptote story. Three regimes, each tested directly on BC Calc 12: (i) if $\deg(\text{top}) < \deg(\text{bot})$, the limit at $\pm \infty$ is $0$ (asymptote $y = 0$); (ii) if $\deg(\text{top}) = \deg(\text{bot})$, the limit is the ratio of leading coefficients (asymptote $y = a_{n} / b_{n}$, here $3/2$); (iii) if $\deg(\text{top}) > \deg(\text{bot})$, the limit diverges to $\pm \infty$ (no horizontal asymptote, possibly a slant asymptote). The Study Guide $\S 3$ tables this. The dominant-term mental move is faster than long division and is the BC-marker-preferred method.
PART II  ·  EXTENDED RESPONSE · SOLUTIONSAP-feeder FRQ + honors · 35 marks

Section B · Worked Solutions

Q6MEDIUMHonors 🇺🇸 US 🇨🇦 ON AP-feeder FRQ §4 Derivative from Definition · AP Calc AB Unit 2 / MCV4U Strand B [8 marks]

$f(x) = x^{2} - 4 x + 5$. (a) Definition. (b) Difference quotient. (c) Take the limit. (d) Verify via power rule + interpret $f'(3)$.

Answer:  (a) $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$  ·  (b) $\dfrac{f(x+h) - f(x)}{h} = 2 x - 4 + h$  ·  (c) $f'(x) = 2 x - 4$  ·  (d) $f'(3) = 2$; slope of tangent at $(3, 2)$ is $2$.

(a) State the definition A1

$$ f'(x) \;=\; \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}. $$ (MCV4U Strand B markers will deduct an A1 if this definition is not written before the algebra begins.)

(b) Compute $f(x + h)$, form the difference quotient, simplify M1·A1·A1

$$ f(x + h) \;=\; (x + h)^{2} - 4 (x + h) + 5 \;=\; x^{2} + 2 x h + h^{2} - 4 x - 4 h + 5. $$ Subtract $f(x) = x^{2} - 4 x + 5$: $$ f(x + h) - f(x) \;=\; (2 x h + h^{2}) + (-4 h) \;=\; h(2 x + h - 4). $$ Divide by $h$ (legal because $h \ne 0$ inside the limit): $$ \frac{f(x + h) - f(x)}{h} \;=\; 2 x + h - 4. $$

(c) Take $h \to 0$ M1·A1

$$ f'(x) \;=\; \lim_{h \to 0} (2 x + h - 4) \;=\; 2 x + 0 - 4 \;=\; 2 x - 4. $$

(d) Verify via power rule + interpret A1·A1

Power rule applied directly: $f(x) = x^{2} - 4 x + 5 \;\Rightarrow\; f'(x) = 2 x - 4 + 0 = 2 x - 4$. $\checkmark$ Matches part (c).
Evaluate at $x = 3$: $f'(3) = 2 (3) - 4 = 2$. Interpretation: the slope of the tangent line to $y = f(x)$ at the point $(3, f(3)) = (3, 2)$ is $2$ units rise per unit run.
"From first principles" earns its method marks from the definition line and the $h$-cancellation, not from the answer. Both AP Calc AB and MCV4U force students to derive the derivative from the limit definition at least once before unleashing the power rule. The grading rubric on these items is structural: an A1 for writing the definition, an M1 for setting up $f(x + h)$ correctly, an M1 for factoring $h$ from the numerator so it cancels with the denominator, an A1 for the simplified difference quotient, and an A1 for the final limit. Students who skip to "$f'(x) = 2 x - 4$ by the power rule" earn zero on parts (a)–(c) even though the final answer is correct. The Study Guide $\S 4$ walks this rubric step-by-step.
Q7MEDIUM 🇨🇦 ON ON Provincial-style §5 Power Rule & Linearity · MCV4U Strand B [8 marks]

(a) $p(x) = 6 x^{4} - 2 x^{3} + 9 x - 14$. (b) $q(x) = 5/x^{2} + 8 \sqrt{x}$. (c) Tangent line to $p$ at $x = 1$.

Answer:  (a) $p'(x) = 24 x^{3} - 6 x^{2} + 9$  ·  (b) $q'(x) = -\dfrac{10}{x^{3}} + \dfrac{4}{\sqrt{x}}$  ·  (c) $y = 27 x - 28$

(a) Power + sum-difference + constant-multiple rules M1·A1

Apply $(c x^{n})' = c n x^{n - 1}$ term-by-term: $$ p'(x) \;=\; 6 \cdot 4 x^{3} - 2 \cdot 3 x^{2} + 9 \cdot 1 - 0 \;=\; 24 x^{3} - 6 x^{2} + 9. $$

(b) Rewrite with rational exponents, then power rule M1·A1·A1

Rewrite: $\dfrac{5}{x^{2}} = 5 x^{-2}$ and $8 \sqrt{x} = 8 x^{1/2}$. So $q(x) = 5 x^{-2} + 8 x^{1/2}$. Differentiate: $$ q'(x) \;=\; 5 \cdot (-2) x^{-3} + 8 \cdot \tfrac{1}{2} x^{-1/2} \;=\; -10 x^{-3} + 4 x^{-1/2} \;=\; -\frac{10}{x^{3}} + \frac{4}{\sqrt{x}}. $$ Note the power rule extends to negative and fractional exponents (MCV4U Strand B and Study Guide $\S 5b$).

(c) Tangent line at $x = 1$ M1·A1·A1

Slope: $m = p'(1) = 24 (1)^{3} - 6 (1)^{2} + 9 = 24 - 6 + 9 = 27$.
Point: $p(1) = 6 - 2 + 9 - 14 = -1$, so $(x_{0}, y_{0}) = (1, -1)$.
Point-slope form: $y - (-1) = 27 (x - 1) \;\Rightarrow\; y = 27 x - 27 - 1 = 27 x - 28$.
Rewrite first, differentiate second: rational-exponent form is the bridge between algebra and calculus. The single most common MCV4U Strand B error on $q(x) = 5/x^{2} + 8 \sqrt{x}$ is to try to differentiate the original expression directly without first rewriting it in $x^{n}$ form. The power rule $\big((x^{n})' = n x^{n - 1}\big)$ only fires when the function is visibly $x^{n}$. The rewrite is two lines, costs nothing, and unlocks the rule. The Study Guide $\S 5b$ frames this as the "negative and fractional exponents" extension to the power rule. A bonus: after differentiating, ON markers want the answer in positive-exponent form with radicals restored, the rewrite trick has to run forward and backward.
Q8HARD 🇨🇦 BC 🇨🇦 AB BC Provincial-style §2 Algebraic Limits · BC Calc 12 / AB Math 31 [8 marks]

(a) $\lim_{x \to 4} \dfrac{x^{2} - 16}{x^{2} - x - 12}$. (b) $\lim_{x \to 0} \dfrac{\sqrt{x + 9} - 3}{x}$. (c) $\lim_{x \to 2} \dfrac{x^{3} - 8}{x - 2}$.

Answer:  (a) $\dfrac{8}{7}$  ·  (b) $\dfrac{1}{6}$  ·  (c) $12$

(a) Factor and cancel M1·A1·A1

Direct substitution gives $\dfrac{16 - 16}{16 - 4 - 12} = \dfrac{0}{0}$, indeterminate. Factor: $$ \frac{x^{2} - 16}{x^{2} - x - 12} \;=\; \frac{(x - 4)(x + 4)}{(x - 4)(x + 3)} \;=\; \frac{x + 4}{x + 3} \quad (x \ne 4). $$ Substitute: $\displaystyle\lim_{x \to 4} \frac{x + 4}{x + 3} = \frac{8}{7}$.

(b) Conjugate technique M1·A1·A1

Direct substitution gives $\dfrac{\sqrt{9} - 3}{0} = \dfrac{0}{0}$. Multiply top and bottom by the conjugate of the numerator, $\sqrt{x + 9} + 3$: $$ \frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} \;=\; \frac{(x + 9) - 9}{x \, (\sqrt{x + 9} + 3)} \;=\; \frac{x}{x \, (\sqrt{x + 9} + 3)} \;=\; \frac{1}{\sqrt{x + 9} + 3} \quad (x \ne 0). $$ Substitute: $\displaystyle\lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$.

(c) Difference-of-cubes pattern M1·A1

Direct substitution gives $\dfrac{8 - 8}{0} = \dfrac{0}{0}$. Use $a^{3} - b^{3} = (a - b)(a^{2} + a b + b^{2})$ with $a = x, b = 2$: $$ \frac{x^{3} - 8}{x - 2} \;=\; \frac{(x - 2)(x^{2} + 2 x + 4)}{x - 2} \;=\; x^{2} + 2 x + 4 \quad (x \ne 2). $$ Substitute: $\displaystyle\lim_{x \to 2} (x^{2} + 2 x + 4) = 4 + 4 + 4 = 12$.
Three algebraic moves, one decision tree. When direct substitution returns $0/0$, the Study Guide $\S 2$ checklist runs: (1) Can I factor a common term out of both numerator and denominator? Try this first, especially on differences of squares, differences of cubes, and quadratic factorisations. (2) Does the indeterminate form involve a square root? Multiply by the conjugate to rationalise. (3) Is there a complex fraction? Combine the inner fractions first, then factor. The same three moves recur on every BC Calc 12 / AB Math 31 / AP Calc AB Unit 1 algebraic-limits problem. Pattern-matching beats brute force: spotting the difference-of-cubes pattern in (c) collapses a six-line problem into one factoring step.
Q9HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §3 Continuity at a Parameter · AP Calc AB Unit 1 / BC Calc 12 [11 marks]

$h(x) = (x^{2} - a^{2})/(x - a)$ for $x \ne a$; $h(a) = b$. (a) Simplify for $x \ne a$. (b) $\lim_{x \to a} h$. (c) Three continuity conditions + $b$. (d) Classify discontinuity at $a = 5$. (e) Sketch + slope.

Answer:  (a) $h(x) = x + a$ for $x \ne a$  ·  (b) $\lim_{x \to a} h(x) = 2a$  ·  (c) $b = 2 a$  ·  (d) at $a = 5$, $b = 10$, the original "hole" is removable (now removed)  ·  (e) line $y = x + 5$ with slope $1$, point $(5, 10)$ filled in.

(a) Factor and cancel M1·A1

Difference of squares: $x^{2} - a^{2} = (x - a)(x + a)$. For $x \ne a$: $$ h(x) \;=\; \frac{(x - a)(x + a)}{x - a} \;=\; x + a. $$

(b) Limit as $x \to a$ M1·A1

Since $h(x) = x + a$ on a punctured neighbourhood of $a$, and $x + a$ is a polynomial (continuous everywhere), $$ \lim_{x \to a} h(x) \;=\; \lim_{x \to a} (x + a) \;=\; 2 a. $$

(c) Three continuity conditions; solve for $b$ M1·A1·A1

The three conditions for continuity of $h$ at $x = a$ are: (i) $h(a)$ is defined; (ii) $\lim_{x \to a} h(x)$ exists; (iii) $\lim_{x \to a} h(x) = h(a)$.
Here $h(a) = b$ (defined for every real $b$), so (i) is automatic. From (b), the limit exists and equals $2 a$, so (ii) holds. Condition (iii) becomes $$ b \;=\; h(a) \;=\; \lim_{x \to a} h(x) \;=\; 2 a, $$ so $b = 2 a$ is exactly the value that makes $h$ continuous at $a$.

(d) Classify at $a = 5$ A1·R1

With $a = 5$, the formula on $x \ne 5$ is $h(x) = x + 5$, and the limit at $x = 5$ is $2 \cdot 5 = 10$. Choosing $b = 10$ via part (c) makes $h(5) = 10 = \lim_{x \to 5} h(x)$, so condition (iii) now holds and $h$ is continuous at $x = 5$. The original discontinuity (when $b$ was unspecified or wrong) was removable: a single point patches it. With $b = 10$, the discontinuity is removed entirely. By contrast, a jump or infinite discontinuity cannot be patched by redefining a single value.

(e) Sketch and slope A1·A1

With $a = 5, b = 10$, the graph of $h$ is the straight line $y = x + 5$ everywhere (including $x = 5$ where it now passes through $(5, 10)$ cleanly, no hole). Slope: $\;m = 1$ (coefficient of $x$). Sketch features: a $45^{\circ}$-ish line crossing the $y$-axis at $(0, 5)$, climbing through $(5, 10)$ with the point filled in, and continuing through $(10, 15)$. No open circle, no jump.
"Removable" means one point repairs the function; "essential" means no single value can. The Study Guide $\S 3$ distinguishes the two by checking whether $\lim_{x \to c} f(x)$ exists as a finite number. If it does, the discontinuity is removable: set $f(c)$ equal to the limit and continuity holds. If the limit fails to exist (jump, oscillation, or blow-up to $\infty$), no single redefinition can repair it. AP Calc AB Unit 1 returns to this distinction relentlessly: every "find $b$ to make $f$ continuous" problem is really a "find the limit" problem in disguise. The signal is the $0/0$ form at $x = c$, which on a real-coefficient rational function always points to a removable hole that algebraic cancellation reveals.
PART III  ·  MODELING / APPLIED · SOLUTIONSUniversal · 28 marks

Section C · Modeling and Applications

Q10MEDIUM 🇨🇦 ON ON Provincial-style §4 Tangent Line & Instantaneous Rate · MCV4U Strand A & B [9 marks]

Drone: $s(t) = -5 t^{2} + 40 t$ (m, s), $0 \le t \le 8$. (a) Average rate on $[1, 3]$. (b) $s'(t)$ + $s'(2)$ in context. (c) Time $s' = 0$ + peak meaning. (d) Tangent at $t = 1$.

Answer:  (a) $\bar v_{[1,3]} = 20$ m/s  ·  (b) $s'(t) = -10 t + 40$; $s'(2) = 20$ m/s upward  ·  (c) $t = 4$ s, peak height $80$ m  ·  (d) $y = 30 t + 5$

(a) Average rate of change on $[1, 3]$ M1·A1

Average rate is the secant slope: $$ \bar v_{[1,3]} \;=\; \frac{s(3) - s(1)}{3 - 1} \;=\; \frac{(-45 + 120) - (-5 + 40)}{2} \;=\; \frac{75 - 35}{2} \;=\; \frac{40}{2} \;=\; 20 \;\text{m/s}. $$

(b) Instantaneous rate via power rule M1·A1

$s'(t) = -5 \cdot 2 t + 40 \cdot 1 = -10 t + 40$. At $t = 2$: $\;s'(2) = -20 + 40 = 20$ m/s. Interpretation: at $t = 2$ seconds, the drone is rising at an instantaneous vertical velocity of $20$ metres per second (positive means upward).
Sanity check: the average rate on $[1, 3]$ also came out to $20$ m/s; by symmetry of the secant about $t = 2$, this match is expected for a quadratic, MCV4U Strand A actually uses this as a discovery exercise.

(c) Time of zero velocity = peak height M1·A1

Solve $s'(t) = 0$: $-10 t + 40 = 0 \Rightarrow t = 4$ s. Interpretation: at $t = 4$ the drone's vertical velocity is momentarily zero, this is the peak of its trajectory before it starts descending. Peak height: $s(4) = -80 + 160 = 80$ m.

(d) Tangent line at $t = 1$ M1·A1·A1

Slope: $s'(1) = -10 + 40 = 30$.
Point: $s(1) = -5 + 40 = 35$, so $(1, 35)$.
Point-slope: $y - 35 = 30 (t - 1) \Rightarrow y = 30 t - 30 + 35 = 30 t + 5$.
Average rate is a secant slope; instantaneous rate is a tangent slope; the limit ties them together. MCV4U Strand A is built on exactly this language: as the interval $[t_{0}, t_{0} + \Delta t]$ shrinks ($\Delta t \to 0$), the secant slope tends to the tangent slope, which is the instantaneous rate of change $s'(t_{0})$. The Study Guide $\S 4$ derivation is verbatim this story: the difference quotient is the average rate; its $h \to 0$ limit is the derivative. In every applied-motion problem, "average velocity over an interval" is a secant calculation; "velocity at a moment" is a derivative calculation; "the rocket is at its peak" is the condition $s' = 0$; "tangent line at a moment" is point-plus-derivative-slope. Provincial markers test each of these four moves in turn.
Q11MEDIUMHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §6 Antiderivative & Indefinite Integral · AP Calc AB Unit 6 [10 marks]

$v(t) = 3 t^{2} - 4 t + 2$ m/s, $s(0) = 5$ m. (a) General antiderivative. (b) Particular antiderivative via $s(0)$. (c) Verify $s' = v$. (d) $\int (4 x^{3} - 6/x^{2} + 5) dx$.

Answer:  (a) $s(t) = t^{3} - 2 t^{2} + 2 t + C$  ·  (b) $s(t) = t^{3} - 2 t^{2} + 2 t + 5$  ·  (c) $s'(t) = 3 t^{2} - 4 t + 2 = v(t)$ $\checkmark$  ·  (d) $x^{4} + \dfrac{6}{x} + 5 x + C$

(a) General antiderivative via reverse power rule M1·A1·A1

Reverse power rule: $\displaystyle\int t^{n} \, dt = \frac{t^{n + 1}}{n + 1} + C$ for $n \ne -1$. Apply term by term: $$ s(t) \;=\; \int (3 t^{2} - 4 t + 2) \, dt \;=\; \frac{3 t^{3}}{3} - \frac{4 t^{2}}{2} + 2 t + C \;=\; t^{3} - 2 t^{2} + 2 t + C. $$ The constant $C$ is mandatory: any constant differentiates to zero, so the antiderivative is determined only up to an additive constant.

(b) Use $s(0) = 5$ to pin down $C$ M1·A1

$s(0) = 0 - 0 + 0 + C = C = 5$. So $C = 5$ and $$ s(t) \;=\; t^{3} - 2 t^{2} + 2 t + 5. $$

(c) Verify $s'(t) = v(t)$ M1·A1

Power rule on $s(t) = t^{3} - 2 t^{2} + 2 t + 5$: $$ s'(t) \;=\; 3 t^{2} - 4 t + 2 + 0 \;=\; v(t). \;\checkmark $$ The "$+ 5$" disappears (derivative of a constant is zero), confirming the rule that any particular antiderivative differentiates back to the original integrand.

(d) Indefinite integral with negative-exponent term M1·A1·A1

Rewrite $\dfrac{1}{x^{2}} = x^{-2}$. Apply reverse power rule: $$ \int 4 x^{3} \, dx \;=\; \frac{4 x^{4}}{4} \;=\; x^{4}, \quad \int -6 x^{-2} \, dx \;=\; -6 \cdot \frac{x^{-1}}{-1} \;=\; 6 x^{-1} \;=\; \frac{6}{x}, \quad \int 5 \, dx \;=\; 5 x. $$ Combine and add $+ C$: $$ \int \left( 4 x^{3} - \frac{6}{x^{2}} + 5 \right) dx \;=\; x^{4} + \frac{6}{x} + 5 x + C. $$ Audit by differentiation: $\;(x^{4})' = 4 x^{3}$; $\;(6/x)' = (6 x^{-1})' = -6 x^{-2} = -6/x^{2}$; $\;(5 x)' = 5$; $\;C' = 0$. Sum: $4 x^{3} - 6/x^{2} + 5$. $\checkmark$
The $+ C$ is its own A1 mark; the "verify by differentiating" step is your free insurance policy. AP Calc AB Unit 6 rubrics universally reserve one A1 for the constant of integration on every indefinite-integral free-response question. Students who omit $+ C$ lose that mark, period; there is no partial credit for "I knew it was there". The Study Guide $\S 6$ also stresses the verification habit: after writing any antiderivative, differentiate it and check that you recover the integrand. The reverse-power-rule formula $\int x^{n} \, dx = x^{n + 1}/(n + 1) + C$ also fails at $n = -1$ (where the answer is $\ln|x| + C$, not in this unit); the negative-exponent term $-6/x^{2} = -6 x^{-2}$ has $n = -2 \ne -1$, so the formula works as written.
Q12HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §7 Definite Integral as Signed Area · AP Calc AB Unit 6 / FTC [9 marks]

$f(x) = 4 - x^{2}$ on $[-2, 3]$. (a) Sketch + intercepts + shading. (b) Sign of area on $[-2, 2]$ vs $[2, 3]$. (c) $\int_{-2}^{2}$ via FTC. (d) $\int_{2}^{3}$ via FTC. (e) Combine; net vs total geometric area.

Answer:  (a) downward parabola, intercepts $x = \pm 2$, vertex $(0, 4)$  ·  (b) $[-2, 2]$ above axis, $+$; $[2, 3]$ below, $-$  ·  (c) $\dfrac{32}{3}$  ·  (d) $-\dfrac{7}{3}$  ·  (e) $\dfrac{25}{3}$, this is net signed area, not total geometric area.

(a) Sketch A1

$f(x) = 4 - x^{2}$ is a downward-opening parabola (leading coefficient $-1$) with vertex at $(0, 4)$ and $x$-intercepts where $4 - x^{2} = 0$, i.e. $x = \pm 2$. On the interval $[-2, 3]$ the graph rises from $(-2, 0)$ up to the vertex $(0, 4)$, then descends through $(2, 0)$ and continues to $(3, -5)$. Shade the region bounded above by the curve and bounded below by the $x$-axis for $x \in [-2, 2]$, and the region bounded above by the $x$-axis and below by the curve for $x \in [2, 3]$ (the curve dips below the axis on this sub-interval).

(b) Sign on each sub-interval A1·A1

On $[-2, 2]$: $f(x) \ge 0$ (the parabola sits above the $x$-axis there), so the area between the curve and the axis is counted positively.
On $[2, 3]$: $f(x) \le 0$ (the parabola dips below the $x$-axis), so the "area" between the curve and the axis is counted negatively by the signed-area convention of the definite integral.

(c) FTC on $[-2, 2]$ M1·A1·A1

Antiderivative: $F(x) = 4 x - \dfrac{x^{3}}{3}$ (drop the $+ C$, it cancels in the FTC subtraction). Then $$ \int_{-2}^{2} (4 - x^{2}) \, dx \;=\; F(2) - F(-2) \;=\; \left( 8 - \tfrac{8}{3} \right) - \left( -8 + \tfrac{8}{3} \right) \;=\; 16 - \tfrac{16}{3} \;=\; \tfrac{48 - 16}{3} \;=\; \tfrac{32}{3}. $$ Positive, consistent with the curve being above the axis on $[-2, 2]$.

(d) FTC on $[2, 3]$ A1

Using the same $F$: $$ \int_{2}^{3} (4 - x^{2}) \, dx \;=\; F(3) - F(2) \;=\; \left( 12 - 9 \right) - \left( 8 - \tfrac{8}{3} \right) \;=\; 3 - 8 + \tfrac{8}{3} \;=\; -5 + \tfrac{8}{3} \;=\; \tfrac{-15 + 8}{3} \;=\; -\tfrac{7}{3}. $$ Negative, consistent with the curve being below the axis on $[2, 3]$, confirming part (b).

(e) Combine; "net signed" vs "total geometric" M1·A1

By the additivity property of the definite integral over adjacent intervals, $$ \int_{-2}^{3} (4 - x^{2}) \, dx \;=\; \int_{-2}^{2} (4 - x^{2}) \, dx + \int_{2}^{3} (4 - x^{2}) \, dx \;=\; \tfrac{32}{3} + \left( -\tfrac{7}{3} \right) \;=\; \tfrac{25}{3}. $$ The result $\dfrac{25}{3}$ is the net signed area on $[-2, 3]$: areas above the axis count positively, areas below count negatively, and the answer is the algebraic sum. Total geometric area (treating both as positive) would instead be $\dfrac{32}{3} + \dfrac{7}{3} = \dfrac{39}{3} = 13$, which requires integrating $|f(x)|$, not $f(x)$.
The definite integral is signed area; "total area between curve and axis" is a different problem. The Study Guide $\S 7$ makes this distinction central. The Fundamental Theorem of Calculus says $\int_{a}^{b} f \, dx = F(b) - F(a)$ for any antiderivative $F$ of $f$, and that quantity is signed: contributions where $f > 0$ add, contributions where $f < 0$ subtract. If a question asks for "the area enclosed between $y = f(x)$ and the $x$-axis on $[a, b]$", you must split at every sign change, integrate each sub-interval separately, take absolute values, then add. AP Calc AB Unit 6 distinguishes the two by language: "definite integral" $=$ signed; "area between" $=$ total geometric (positive). The signed result $\dfrac{25}{3}$ and the geometric result $13$ disagree here precisely because the curve crosses the axis at $x = 2$. The constant $+ C$ in the antiderivative cancels in $F(b) - F(a)$, so it drops out of the FTC computation cleanly, a nice symmetry with the $+ C$-mandatory indefinite case.