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

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）：

$t$	0	2	5	8	12
$T(t)$	72	75	80	83	88

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

$$\begin{aligned} T'(5) &\approx \frac{T(8) - T(2)}{8 - 2} \\[4pt] &= \frac{83 - 75}{6} \\[4pt] &= \frac{8}{6} \approx 1.333 \text{ °F/min} \end{aligned}$$

A graph of $g$ shows an open circle at $(2, 4)$, a filled dot at $(2, 1)$, and the curve rising smoothly toward height $4$ from both sides of $x = 2$. Find $\lim_{x \to 2} g(x)$ and $g(2)$.函数 $g$ 的图像在 $(2, 4)$ 处为空心圆，在 $(2, 1)$ 处为实心点，且曲线从 $x = 2$ 两侧都平滑地趋向高度 $4$。求 $\lim_{x \to 2} g(x)$ 与 $g(2)$。

Step 1 — Read each one-sided limit. As $x \to 2^-$ and $x \to 2^+$, the curve approaches height $4$ from both sides.第 1 步——读取每个单侧极限。当 $x \to 2^-$ 与 $x \to 2^+$ 时，曲线从两侧都趋向高度 $4$。

$$\lim_{x \to 2^-} g(x) = 4 \qquad \lim_{x \to 2^+} g(x) = 4$$

Step 2 — Combine. The one-sided limits agree, so the two-sided limit exists and equals $4$.第 2 步——合并。两个单侧极限相等，故双侧极限存在且等于 $4$。

$$\lim_{x \to 2} g(x) = 4$$

Step 3 — Read the value separately. The filled dot gives $g(2) = 1$. The limit ($4$) is not the function value ($1$): the graph has a removable discontinuity at $x = 2$.第 3 步——单独读取函数值。实心点给出 $g(2) = 1$。极限（$4$）不等于函数值（$1$）：图像在 $x = 2$ 处为可去间断点。

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

$x$	1.9	1.99	1.999	2.001	2.01	2.1
$f(x)$	3.9	3.99	3.999	4.001	4.01	4.1

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

$$\therefore \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$

A table for $f$ near $x = 0$ shows the values below. Estimate $\lim_{x \to 0} f(x)$.$f$ 在 $x = 0$ 附近的表格取值如下。估计 $\lim_{x \to 0} f(x)$。

$x$	-0.1	-0.01	0.01	0.1
$f(x)$	-1.98	-1.999	3.001	3.02

Step 1 — Read each side. From the left the values approach $-2$; from the right they approach $3$.第 1 步——分别读取两侧。从左侧取值趋于 $-2$；从右侧取值趋于 $3$。

$$\lim_{x \to 0^-} f(x) \approx -2 \qquad \lim_{x \to 0^+} f(x) \approx 3$$

Step 2 — Compare. The one-sided estimates disagree ($-2 \neq 3$), so the two-sided limit does not exist. A table is enough to reveal this — you do not need a formula.第 2 步——比较。两侧的估计值不一致（$-2 \neq 3$），故双侧极限不存在。仅凭表格即可看出这一点——不需要公式。

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：

$$\begin{aligned} &= 2(3)^2 + 5(3) - 1 \\ &= 2(9) + 15 - 1 \\ &= 18 + 15 - 1 \\ &= 32 \end{aligned}$$

Find $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} + 3}{x - 1}$ using the limit laws, naming each law you use.利用极限法则求 $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} + 3}{x - 1}$，并指出每一步所用的法则。

Step 1 — Check the denominator. At $x = 4$, the denominator is $4 - 1 = 3 \neq 0$, so the quotient law applies (no indeterminate form).第 1 步——检查分母。当 $x = 4$ 时，分母为 $4 - 1 = 3 \neq 0$，故可用商法则（无不定式）。

Step 2 — Evaluate numerator and denominator separately.第 2 步——分别求分子与分母的极限。

$$\begin{aligned} \lim_{x \to 4}(\sqrt{x} + 3) &= \sqrt{4} + 3 = 5 \qquad \text{(sum and root laws)} \\[4pt] \lim_{x \to 4}(x - 1) &= 4 - 1 = 3 \qquad \text{(difference law)} \end{aligned}$$

Step 3 — Apply the quotient law.第 3 步——应用商法则。

$$\lim_{x \to 4} \frac{\sqrt{x} + 3}{x - 1} = \frac{5}{3}$$

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：

$$\begin{aligned} &= \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} \\[4pt] &= \lim_{x \to 3} (x+3) \qquad\text{(cancel } x-3\text{)} \\[4pt] &= 3 + 3 = 6 \end{aligned}$$

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：

$$\begin{aligned} &= \lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{(x+4)-4}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} \\[4pt] &= \lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} \\[4pt] &= \frac{1}{\sqrt{4}+2} = \frac{1}{4} \end{aligned}$$

Find $\displaystyle\lim_{x \to 0} \frac{\frac{1}{x + 3} - \frac{1}{3}}{x}$.求 $\displaystyle\lim_{x \to 0} \frac{\frac{1}{x + 3} - \frac{1}{3}}{x}$。

Step 1 — Identify the form. Direct substitution gives $\frac{0}{0}$, so simplify first.第 1 步——判别形式。直接代入得 $\frac{0}{0}$，故先化简。

Step 2 — Combine the top fractions over the common denominator $3(x+3)$:第 2 步——将分子两项通分，公分母为 $3(x+3)$：

$$\frac{1}{x+3} - \frac{1}{3} = \frac{3 - (x+3)}{3(x+3)} = \frac{-x}{3(x+3)}$$

Step 3 — Divide by $x$ (multiply by $\frac{1}{x}$) and cancel:第 3 步——除以 $x$（即乘 $\frac{1}{x}$）并约分：

$$\frac{-x}{3(x+3)} \cdot \frac{1}{x} = \frac{-1}{3(x+3)}$$

Step 4 — Substitute.第 4 步——代入。

$$\lim_{x \to 0} \frac{-1}{3(x+3)} = \frac{-1}{3(3)} = -\frac{1}{9}$$

Determine $\displaystyle\lim_{x \to 3} \frac{x + 2}{x - 3}$.求 $\displaystyle\lim_{x \to 3} \frac{x + 2}{x - 3}$。

Step 1 — Substitute to classify the form. The numerator $\to 5$ and the denominator $\to 0$, giving $\frac{5}{0}$ — a nonzero-over-zero form. This is not $\frac{0}{0}$, so do not factor; the limit is infinite or DNE.第 1 步——代入以判别形式。分子 $\to 5$，分母 $\to 0$，得 $\frac{5}{0}$——非零比零型。它不是 $\frac{0}{0}$，故不应因式分解；极限为无穷或不存在。

Step 2 — Test each side with a sign analysis.第 2 步——用符号分析检验两侧。

$$\lim_{x \to 3^-} \frac{x+2}{x-3} = \frac{5}{0^-} = -\infty \qquad \lim_{x \to 3^+} \frac{x+2}{x-3} = \frac{5}{0^+} = +\infty$$

Step 3 — Conclude. The one-sided limits disagree ($-\infty$ vs $+\infty$), so the two-sided limit does not exist, and $x = 3$ is a vertical asymptote.第 3 步——下结论。两个单侧极限不一致（$-\infty$ 与 $+\infty$），故双侧极限不存在，且 $x = 3$ 为垂直渐近线。

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：

$$\lim_{x \to 0} x^2 \sin\!\left(\tfrac{1}{x}\right) = 0$$

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 $\frac{\sin u}{u} \to 1$：

$$\begin{aligned} &= \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3x}{5x} \\[4pt] &= \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3}{5} \\[4pt] &= 1 \cdot \frac{3}{5} = \frac{3}{5} \end{aligned}$$

Find $\displaystyle\lim_{x \to 0} x\cos\!\left(\frac{1}{x^2}\right)$.求 $\displaystyle\lim_{x \to 0} x\cos\!\left(\frac{1}{x^2}\right)$。

Step 1 — Bound the oscillating factor. For all $x \neq 0$, $-1 \le \cos\!\left(\frac{1}{x^2}\right) \le 1$.第 1 步——对振荡因子定界。对所有 $x \neq 0$，$-1 \le \cos\!\left(\frac{1}{x^2}\right) \le 1$。

Step 2 — Multiply by $|x|$. Since $|x| \ge 0$, multiplying preserves the inequality:第 2 步——乘以 $|x|$。因 $|x| \ge 0$，相乘保持不等式方向：

$$-|x| \;\le\; x\cos\!\left(\frac{1}{x^2}\right) \;\le\; |x|$$

Step 3 — Squeeze. Both $\lim_{x \to 0}(-|x|) = 0$ and $\lim_{x \to 0}|x| = 0$, so by the Squeeze Theorem the middle limit is $0$.第 3 步——夹逼。由于 $\lim_{x \to 0}(-|x|) = 0$ 且 $\lim_{x \to 0}|x| = 0$，由夹逼定理，中间的极限为 $0$。

$$\lim_{x \to 0} x\cos\!\left(\frac{1}{x^2}\right) = 0$$

Classify the discontinuity of $f$ at $x = 1$, where $f(x) = \begin{cases} 2x, & x < 1 \\ x + 3, & x \ge 1 \end{cases}$.判定 $f$ 在 $x = 1$ 处的不连续类型，其中 $f(x) = \begin{cases} 2x, & x < 1 \\ x + 3, & x \ge 1 \end{cases}$。

Step 1 — One-sided limits.第 1 步——单侧极限。

$$\lim_{x \to 1^-} 2x = 2 \qquad \lim_{x \to 1^+} (x + 3) = 4$$

Step 2 — Compare. Both one-sided limits are finite but $2 \neq 4$, so the two-sided limit does not exist. Because the two sides are finite and simply "jump," this is a jump discontinuity.第 2 步——比较。两个单侧极限均有限，但 $2 \neq 4$，故双侧极限不存在。由于两侧都有限且发生"跳跃"，这是跳跃间断点。

Classify the discontinuity of $f(x) = \dfrac{x + 1}{x - 2}$ at $x = 2$.判定 $f(x) = \dfrac{x + 1}{x - 2}$ 在 $x = 2$ 处的不连续类型。

Step 1 — Check the form. At $x = 2$: numerator $\to 3$, denominator $\to 0$, giving $\frac{3}{0}$ — nonzero over zero, so the factor does not cancel.第 1 步——检查形式。当 $x = 2$ 时：分子 $\to 3$，分母 $\to 0$，得 $\frac{3}{0}$——非零比零，因式不能约去。

Step 2 — One-sided behavior. As $x \to 2^-$, $f \to -\infty$; as $x \to 2^+$, $f \to +\infty$. At least one side is infinite, so this is an infinite (essential) discontinuity with a vertical asymptote at $x = 2$.第 2 步——单侧行为。当 $x \to 2^-$ 时 $f \to -\infty$；当 $x \to 2^+$ 时 $f \to +\infty$。至少一侧为无穷，故为无穷（本性）间断点，在 $x = 2$ 处有垂直渐近线。

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）。

Show that $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 6, & x = 3 \end{cases}$ is continuous at $x = 3$.证明 $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 6, & x = 3 \end{cases}$ 在 $x = 3$ 处连续。

Condition 1 — $f(3)$ is defined. By the rule, $f(3) = 6$. ✓条件 1——$f(3)$ 有定义。由定义，$f(3) = 6$。✓

Condition 2 — the limit exists. For $x \neq 3$, factor and cancel:条件 2——极限存在。当 $x \neq 3$ 时，因式分解并约分：

$$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x + 3) = 6$$

Condition 3 — they match. Since $\lim_{x \to 3} f(x) = 6 = f(3)$, all three conditions hold, so $f$ is continuous at $x = 3$. The piecewise value $6$ was chosen precisely to plug the hole.条件 3——两者相等。由于 $\lim_{x \to 3} f(x) = 6 = f(3)$，三个条件都满足，故 $f$ 在 $x = 3$ 处连续。分段取值 $6$ 恰好是为了补上这个洞。

Show that $f(x) = \sqrt{4 - x^2}$ is continuous on the closed interval $[-2, 2]$.证明 $f(x) = \sqrt{4 - x^2}$ 在闭区间 $[-2, 2]$ 上连续。

Step 1 — Domain check. $f$ is defined exactly where $4 - x^2 \ge 0$, i.e. on $[-2, 2]$, so the interval matches the natural domain.第 1 步——定义域检查。$f$ 恰好在 $4 - x^2 \ge 0$ 处有定义，即 $[-2, 2]$，与所给区间一致。

Step 2 — Interior continuity. $4 - x^2$ is a polynomial, so it is continuous everywhere. Since $\sqrt{u}$ is continuous for every $u \ge 0$, the composition $f(x) = \sqrt{4 - x^2}$ is continuous at every point of the open interval $(-2, 2)$.第 2 步——内部连续性。$4 - x^2$ 是多项式，处处连续。又因 $\sqrt{u}$ 在每个 $u \ge 0$ 处连续，复合函数 $f(x) = \sqrt{4 - x^2}$ 在开区间 $(-2, 2)$ 内每一点都连续。

Step 3 — Endpoint (one-sided) checks. Continuity on a closed interval only requires the appropriate one-sided limit to match the function value at each endpoint:第 3 步——端点（单侧）检查。闭区间上的连续性只要求在每个端点处，相应的单侧极限等于函数值：

$$\lim_{x \to -2^+} \sqrt{4 - x^2} = \sqrt{4 - 4} = 0 = f(-2) \qquad \lim_{x \to 2^-} \sqrt{4 - x^2} = \sqrt{4 - 4} = 0 = f(2)$$

Both one-sided limits agree with the function value, so $f$ is continuous on all of $[-2, 2]$. This same three-step pattern — domain, interior, endpoints — works for any interval-continuity question: don't stop at "the interior is fine," always close the loop at the boundary.两个单侧极限都与函数值相符，故 $f$ 在整个 $[-2, 2]$ 上连续。这个三步套路——定义域、内部、端点——适用于任何区间连续性问题：不要止步于"内部没问题"，一定要在边界处把结论闭合。

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：

$$2k + 1 = 3 \;\;\Longrightarrow\;\; 2k = 2 \;\;\Longrightarrow\;\; k = 1$$

Find the one-sided limits of $f(x) = \dfrac{x - 4}{(x + 1)^2}$ as $x \to -1$.求 $f(x) = \dfrac{x - 4}{(x + 1)^2}$ 当 $x \to -1$ 时的单侧极限。

Step 1 — Form check. At $x = -1$: numerator $\to -5$, denominator $\to 0$, giving $\frac{-5}{0}$ — infinite.第 1 步——形式检查。当 $x = -1$ 时：分子 $\to -5$，分母 $\to 0$，得 $\frac{-5}{0}$——无穷。

Step 2 — Sign of the denominator. Because $(x+1)^2$ is a square, it approaches $0^+$ from both sides.第 2 步——分母的符号。因 $(x+1)^2$ 为平方，从两侧都趋于 $0^+$。

Step 3 — Combine signs. Negative numerator over a tiny positive denominator gives $-\infty$ on each side:第 3 步——综合符号。负的分子除以极小的正分母，两侧都得 $-\infty$：

$$\lim_{x \to -1^-} f(x) = -\infty \qquad \lim_{x \to -1^+} f(x) = -\infty$$

Both sides agree at $-\infty$, so we may also write $\lim_{x \to -1} f(x) = -\infty$, and $x = -1$ is a vertical asymptote.两侧都为 $-\infty$，故也可写 $\lim_{x \to -1} f(x) = -\infty$，且 $x = -1$ 为垂直渐近线。

Find $\displaystyle\lim_{x \to \infty} \frac{6x^2 - 5x + 1}{2x^2 + 7}$.求 $\displaystyle\lim_{x \to \infty} \frac{6x^2 - 5x + 1}{2x^2 + 7}$。

Step 1 — Divide every term by $x^2$ (the highest power in the denominator):第 1 步——每一项都除以 $x^2$（分母中的最高次幂）：

$$\frac{6x^2 - 5x + 1}{2x^2 + 7} = \frac{6 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{7}{x^2}}$$

Step 2 — Send $x \to \infty$. Every $\frac{1}{x}$ and $\frac{1}{x^2}$ term vanishes:第 2 步——令 $x \to \infty$。每个 $\frac{1}{x}$ 与 $\frac{1}{x^2}$ 项都趋于 $0$：

$$\lim_{x \to \infty} \frac{6 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{7}{x^2}} = \frac{6 - 0 + 0}{2 + 0} = 3$$

The ratio of leading coefficients $\frac{6}{2} = 3$ confirms the horizontal asymptote $y = 3$.首项系数之比 $\frac{6}{2} = 3$ 印证了水平渐近线 $y = 3$。

Find $\displaystyle\lim_{x \to \infty} \frac{\sqrt{9x^2 + 1}}{x + 4}$.求 $\displaystyle\lim_{x \to \infty} \frac{\sqrt{9x^2 + 1}}{x + 4}$。

Step 1 — Estimate the top's degree. For large $x$, $\sqrt{9x^2 + 1} \approx \sqrt{9x^2} = 3x$, so the top behaves like a degree-$1$ term, matching the bottom.第 1 步——估计分子的次数。当 $x$ 很大时，$\sqrt{9x^2 + 1} \approx \sqrt{9x^2} = 3x$，故分子表现为一次项，与分母同次。

Step 2 — Divide by $x$. For $x > 0$, $x = \sqrt{x^2}$, so pull it inside the root:第 2 步——除以 $x$。当 $x > 0$ 时 $x = \sqrt{x^2}$，可将其移入根号内：

$$\frac{\sqrt{9x^2 + 1}}{x + 4} = \frac{\sqrt{9 + \frac{1}{x^2}}}{1 + \frac{4}{x}}$$

Step 3 — Take the limit.第 3 步——求极限。

$$\lim_{x \to \infty} \frac{\sqrt{9 + \frac{1}{x^2}}}{1 + \frac{4}{x}} = \frac{\sqrt{9}}{1} = 3$$

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 $f(c) = 0$。✓

Let $f(x) = x^2 + x$. Show that $f$ takes the value $3$ somewhere on $[0, 2]$.设 $f(x) = x^2 + x$。证明 $f$ 在 $[0, 2]$ 上某处取值 $3$。

Step 1 — Continuity. $f$ is a polynomial, hence continuous on $[0, 2]$.第 1 步——连续性。$f$ 是多项式，故在 $[0, 2]$ 上连续。

Step 2 — Endpoints straddle the target $d = 3$.第 2 步——两端点夹住目标值 $d = 3$。

$$f(0) = 0 \qquad f(2) = 4 + 2 = 6, \qquad 0 < 3 < 6$$

Step 3 — Apply the IVT. Since $f$ is continuous on $[0, 2]$ and $3$ lies between $f(0) = 0$ and $f(2) = 6$, there exists $c \in (0, 2)$ with $f(c) = 3$. (Solving $x^2 + x - 3 = 0$ confirms $c = \frac{-1 + \sqrt{13}}{2} \approx 1.30$.)第 3 步——应用介值定理。由于 $f$ 在 $[0, 2]$ 上连续，且 $3$ 位于 $f(0) = 0$ 与 $f(2) = 6$ 之间，故存在 $c \in (0, 2)$ 使 $f(c) = 3$。（解 $x^2 + x - 3 = 0$ 得 $c = \frac{-1 + \sqrt{13}}{2} \approx 1.30$，予以印证。）