Show all supporting work on scratch paper. On the AP Exam, Section I is split into a no-calculator and a calculator-allowed part, each question below is labeled accordingly.请将所有辅助解题过程写在草稿纸上。AP 考试第一部分分为禁用计算器和允许使用计算器两节，下方每道题均已注明。
Q1EASY1.4 Estimating from Tables1.4 表格估算极限No Calculator
The table gives values of $f(x)$ near $x=2$.下表给出了 $f(x)$ 在 $x=2$ 附近的值。
$x$
1.9
1.99
1.999
2.001
2.01
2.1
$f(x)$
4.61
4.9601
4.996
5.004
5.0401
5.41
Based on the table, $\displaystyle\lim_{x\to 2}f(x)$ is best estimated by根据表格，$\displaystyle\lim_{x\to 2}f(x)$ 最佳估计值为
Q6MEDIUM1.3 / 1.8 Squeeze Theorem from a Graph1.3 / 1.8 由图像应用夹逼定理No Calculator
The solid curve is the graph of $g$. The two dashed curves trap it: for every $x$, the lower dashed curve is at most $g(x)$ and the upper dashed curve is at least $g(x)$. From the graph, $\displaystyle\lim_{x\to 0}g(x)=$实线为 $g$ 的图像，两条虚线将其夹住：对每个 $x$，下方虚线不超过 $g(x)$，上方虚线不小于 $g(x)$。由图像可知 $\displaystyle\lim_{x\to 0}g(x)=$
Q8MEDIUM1.11 / 1.13 Continuity at a Point1.11 / 1.13 点连续性No Calculator
Let $f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x\ne 2\\[4pt]k, & x=2\end{cases}$. For what value of $k$ is $f$ continuous at $x=2$?设 $f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x\ne 2\\[4pt]k, & x=2\end{cases}$，$k$ 取何值时 $f$ 在 $x=2$ 处连续？
Let $f$ be continuous on $[0,3]$ with $f(0)=-2$ and $f(3)=5$. Which conclusion does the IVT guarantee?设 $f$ 在 $[0,3]$ 上连续，且 $f(0)=-2$，$f(3)=5$。介值定理可保证哪个结论？
(A) $f(c)=0$ for some $c\in(0,3)$.对某个 $c\in(0,3)$ 成立。
(B) $f(c)=6$ for some $c\in(0,3)$.对某个 $c\in(0,3)$ 成立。
(C) $f(c)=-3$ for some $c\in(0,3)$.对某个 $c\in(0,3)$ 成立。
(D) $f$ is differentiable on $(0,3)$.在 $(0,3)$ 上可导。
The continuous function $h$ has the selected values below. What is the minimum number of real zeros of $h$ on $[1,9]$ guaranteed by the IVT?连续函数 $h$ 的部分值如下表所示。介值定理能保证 $h$ 在 $[1,9]$ 上至少有多少个实零点？
Q20MEDIUM1.10 Discontinuity in Context1.10 情境中的间断No Calculator
A parking garage charges $\$4$ for the first hour or any part of it, and $\$3$ for each additional hour or part of an hour. Let $C(t)$ be the charge in dollars for a stay of $t$ hours, $0 \lt t \le 4$.某停车场首小时（不足一小时按一小时计）收费 $\$4$，其后每增加一小时（不足一小时按一小时计）收费 $\$3$。设 $C(t)$ 为停留 $t$ 小时的费用（美元），$0 \lt t \le 4$。
Which statement about $C$ at $t=2$ is correct?关于 $C$ 在 $t=2$ 处，下列哪个说法正确？
(A)$C$ is continuous at $t=2$, since $C(2)=7$ and the charge is defined there.$C$ 在 $t=2$ 处连续，因为 $C(2)=7$ 且该处有定义。
(B)$\displaystyle\lim_{t\to 2}C(t)$ exists but is not equal to $C(2)$, so the discontinuity is removable.$\displaystyle\lim_{t\to 2}C(t)$ 存在但不等于 $C(2)$，故为可去不连续点。
(C)$C$ has a jump discontinuity at $t=2$, and the jump of $\$3$ is the cost of entering the third hour.$C$ 在 $t=2$ 处为跳跃不连续，跳跃量 $\$3$ 即进入第三小时的费用。
(D)$C$ has an infinite discontinuity at $t=2$.$C$ 在 $t=2$ 处为无穷不连续。
Q21HARD1.6 Comparing Rates of Approach1.6 比较趋近速度No Calculator
Each of the four quantities below approaches $0$ as $x\to 0^{+}$. Without evaluating any limit numerically, order them from smallest to largest for values of $x$ close to $0^{+}$.下列四个量在 $x\to 0^{+}$ 时均趋于 $0$。在不进行数值计算的前提下，将它们按 $x$ 接近 $0^{+}$ 时由小到大排序。
Q22HARD1.5 / 1.6 Solving for a Parameter1.5 / 1.6 求参数No Calculator
Let $k$ be a constant. For which value of $k$ does $\displaystyle\lim_{x\to 3}\dfrac{x^{2}+kx-12}{x-3}$ exist, and what is the limit for that $k$?设 $k$ 为常数。$k$ 取何值时 $\displaystyle\lim_{x\to 3}\dfrac{x^{2}+kx-12}{x-3}$ 存在？此时该极限等于多少？
(A) $k=1$, and the limit is $7$.极限为 $7$。
(B) $k=1$, and the limit is $-4$.极限为 $-4$。
(C) $k=-1$, and the limit is $5$.极限为 $5$。
(D)No value of $k$ makes the limit exist.不存在使该极限存在的 $k$ 值。
Free-response answers must include complete setup: algebraic manipulation, stated theorem conditions for IVT/Squeeze, and interval justification for continuity. Units and contextual explanations are required where indicated.自由解答题须包含完整解题过程：代数化简、介值定理/夹逼定理的条件说明，以及连续性的区间论证。在有要求的情况下，需写出单位和情境说明。
Evaluate the limits in (a) to (c), showing all algebraic steps. Part (d) asks for a justification, not a calculation.求 (a) 至 (c) 各极限，展示完整代数步骤。(d) 要求作出论证，而非计算。
(d)The table gives values of a different function $g$ near $x=2$. A student writes: "the table proves that $\displaystyle\lim_{x\to 2}g(x)=5$." State precisely what the table does establish and what it does not, and give a reason.下表给出另一函数 $g$ 在 $x=2$ 附近的取值。某学生写道："此表证明了 $\displaystyle\lim_{x\to 2}g(x)=5$。"请准确说明该表能确立什么、不能确立什么，并给出理由。
(a)Find $\displaystyle\lim_{x\to 3^-}f(x)$ and explain how this determines a restriction on $a$ and $b$.求 $\displaystyle\lim_{x\to 3^-}f(x)$，并说明这如何对 $a$ 和 $b$ 施加限制。
(b)Determine values of $a$ and $b$ that make $f$ continuous on $\mathbb{R}$. Show the system you solve.求使 $f$ 在 $\mathbb{R}$ 上连续的 $a$ 和 $b$ 的值，写出所建立的方程组。
(c)With the values from (b), classify the discontinuity of $f'$ at $x=3$ and at $x=5$ (if any).利用 (b) 中求得的值，判断 $f'$ 在 $x=3$ 和 $x=5$ 处的不连续类型（若存在）。(Preview of Unit 2: derivative discontinuity. Cram-track students may skip this part.)（第二单元预览，导数的不连续性。备考冲刺学生可跳过此部分。）
Let $f(x)=\dfrac{2x^{2}-x-6}{x^{2}-4}$.设 $f(x)=\dfrac{2x^{2}-x-6}{x^{2}-4}$。
(a)Find all vertical asymptotes of $f$. Justify with one-sided limits.求 $f$ 的所有竖直渐近线，用单侧极限加以论证。
(b)Find any removable discontinuities and state the value needed to remove each.求所有可去不连续点，并说明消除各不连续点所需的函数值。
(c)Find $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to -\infty}f(x)$, and state the horizontal asymptote.求 $\displaystyle\lim_{x\to\infty}f(x)$ 和 $\displaystyle\lim_{x\to -\infty}f(x)$，并写出水平渐近线。
A diver's depth $d(t)$ in meters at time $t$ seconds is continuous on $[0,20]$.潜水员的深度 $d(t)$（单位：米）在时刻 $t$（单位：秒）处连续，定义在 $[0,20]$ 上。
$t$ (s)
0
4
10
15
20
$d(t)$ (m)
0
8
22
18
5
(a)Use the IVT to justify that there is a time in $(0,10)$ when the diver is exactly $15$ meters deep.利用介值定理证明：在 $(0,10)$ 内存在某时刻，潜水员恰好处于 $15$ 米深处。
(b)Is the IVT enough to conclude that the diver is $15$ meters deep at some time in $(10,20)$? Justify.介值定理是否足以得出在 $(10,20)$ 内某时刻潜水员处于 $15$ 米深处的结论？请说明理由。
(c)A student claims the IVT guarantees that the diver's depth equals $25$ meters for some $t\in(0,20)$. Is the claim correct? Explain.某学生声称，介值定理保证在某个 $t\in(0,20)$ 处潜水员的深度等于 $25$ 米。该说法是否正确？请解释。
(d)What is the minimum number of times the diver can be at depth $15$ meters on $[0,20]$? Justify.在 $[0,20]$ 上，潜水员至少有多少次处于 $15$ 米深处？请说明理由。
Let $f(x)=x^{2}\cos\!\bigl(\tfrac{1}{x}\bigr)$ for $x\ne 0$, and define $f(0)=0$.设 $f(x)=x^{2}\cos\!\bigl(\tfrac{1}{x}\bigr)$（$x\ne 0$），并定义 $f(0)=0$。
(a)Show, using the Squeeze Theorem, that $\displaystyle\lim_{x\to 0}f(x)=0$. State the bounding inequalities and verify all conditions.利用夹逼定理证明 $\displaystyle\lim_{x\to 0}f(x)=0$，写出界定不等式并验证所有条件。
(b)Use part (a) to explain why $f$ is continuous at $x=0$.利用 (a) 的结论说明 $f$ 在 $x=0$ 处连续。
(c)Compute $\displaystyle\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}$ using the Squeeze Theorem. Interpret the meaning of the limit.用夹逼定理计算 $\displaystyle\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}$，并解释该极限的含义。
FRQ 6HARDFRQ LEVEL1.11 / 1.16 Piecewise Model in Context1.11 / 1.16 情境中的分段模型No Calculator
A courier charges by parcel weight $w$, measured in kilograms. Let $k$ be a positive constant. The charge $C(w)$, in dollars, is modeled by某快递公司按包裹重量 $w$（单位：千克）计费。设 $k$ 为正常数，费用 $C(w)$（单位：美元）的模型为
$$ C(w)=\begin{cases} 8, & 0 \lt w \le 2\\[4pt] 8+k(w-2), & 2 \lt w \le 10\\[4pt] 40-\dfrac{120}{w}, & 10 \lt w \le 30 \end{cases} $$
(a)Find $\displaystyle\lim_{w\to 2^{-}}C(w)$ and $\displaystyle\lim_{w\to 2^{+}}C(w)$. Determine whether $C$ is continuous at $w=2$, and state whether your conclusion depends on the value of $k$. Justify your answer.求 $\displaystyle\lim_{w\to 2^{-}}C(w)$ 与 $\displaystyle\lim_{w\to 2^{+}}C(w)$。判断 $C$ 在 $w=2$ 处是否连续，并说明该结论是否依赖于 $k$ 的取值。请给出理由。
(b)Find the value of $k$ for which $C$ is continuous at $w=10$. Show the equation you solve.求使 $C$ 在 $w=10$ 处连续的 $k$ 值，写出所建立的方程。
(c)Using the value of $k$ from part (b), interpret the meaning of $k$ in the context of this model. Include units.利用 (b) 中求得的 $k$ 值，结合本模型情境解释 $k$ 的含义，并写明单位。
(d)Using the value of $k$ from part (b), show that there is a parcel weight between $10$ and $30$ kilograms for which the charge is exactly $\$35$. Name the theorem you use and verify each of its hypotheses.利用 (b) 中求得的 $k$ 值，证明在 $10$ 至 $30$ 千克之间存在某一包裹重量，使费用恰为 $\$35$。写出所用定理的名称，并逐条验证其条件。