Unit A1: Limits
and Continuity第 A1 单元：极限
与连续性

Evaluate $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.

Direct substitution gives $0/0$, an indeterminate form, so the algebra must be simplified first. Factor the numerator:

$$ \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x - 3} = x + 3 \quad (x \neq 3). $$

The original function and $x+3$ agree everywhere except at $x = 3$, and limits ignore the single point $x = a$. Therefore

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

Note $f(3)$ is undefined, yet the limit is $6$. The graph has a hole at $(3, 6)$.

求 $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$。

直接代入（direct substitution）得到 $0/0$，这是一个不定式（indeterminate form），所以必须先做代数化简。把分子因式分解：

$$ \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x - 3} = x + 3 \quad (x \neq 3). $$

原函数与 $x+3$ 除了在 $x = 3$ 处之外处处相同，而极限会忽略 $x = a$ 这一单点。因此

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

注意 $f(3)$ 无定义，但极限是 $6$。图像在 $(3, 6)$ 处有一个空洞。

Let $f(x) = \dfrac{|x|}{x}$ and decide whether $\lim_{x \to 0} f(x)$ exists.

For $x > 0$ we have $|x| = x$, so $f(x) = x/x = 1$. For $x < 0$ we have $|x| = -x$, so $f(x) = -x/x = -1$. The two sides therefore approach different targets:

$$ \lim_{x \to 0^+} f(x) = 1, \qquad \lim_{x \to 0^-} f(x) = -1. $$

Because the one-sided limits disagree, the two-sided limit does not exist. This is the prototype of a jump discontinuity: each side is perfectly well behaved on its own, yet they refuse to meet. Note that $f(0)$ is also undefined here, but that is irrelevant; even if we forced a value at $0$, the limit would still fail because the sides disagree.

设 $f(x) = \dfrac{|x|}{x}$，判断 $\lim_{x \to 0} f(x)$ 是否存在。

当 $x > 0$ 时 $|x| = x$，故 $f(x) = x/x = 1$。当 $x < 0$ 时 $|x| = -x$，故 $f(x) = -x/x = -1$。因此两侧趋向不同的目标：

$$ \lim_{x \to 0^+} f(x) = 1, \qquad \lim_{x \to 0^-} f(x) = -1. $$

由于两个单侧极限不一致，双侧极限不存在。这是跳跃间断（jump discontinuity）的典型：每一侧单独看都规规矩矩，却彼此不相接。注意这里 $f(0)$ 同样无定义，但这无关紧要；即使我们强行在 $0$ 处赋一个值，极限仍不存在，因为两侧不一致。

Examine $\lim_{x \to 0} \sin\!\big(\tfrac{1}{x}\big)$.

As $x \to 0$, the input $\tfrac{1}{x}$ runs off to $\pm\infty$, so $\sin(\tfrac1x)$ cycles through its full range $[-1,1]$ infinitely often. Concretely, take two sequences approaching $0$:

$$ x_n = \frac{1}{2\pi n} \ \Rightarrow\ \sin\!\Big(\tfrac{1}{x_n}\Big) = \sin(2\pi n) = 0, \qquad y_n = \frac{1}{2\pi n + \pi/2} \ \Rightarrow\ \sin\!\Big(\tfrac{1}{y_n}\Big) = 1. $$

Both $x_n \to 0$ and $y_n \to 0$, yet the outputs sit at $0$ and at $1$ respectively. A single limit value $L$ would have to absorb both, which is impossible, so the limit does not exist. Contrast this with $\lim_{x\to 0} x\sin(\tfrac1x)$, which does equal $0$: there the wild oscillation is multiplied by $x \to 0$ and the Squeeze Theorem of Section 3 forces convergence.

考察 $\lim_{x \to 0} \sin\!\big(\tfrac{1}{x}\big)$。

当 $x \to 0$ 时，输入 $\tfrac{1}{x}$ 奔向 $\pm\infty$，因此 $\sin(\tfrac1x)$ 在其整个值域 $[-1,1]$ 上无限多次往复循环。具体地，取两个趋于 $0$ 的数列（sequence）：

$$ x_n = \frac{1}{2\pi n} \ \Rightarrow\ \sin\!\Big(\tfrac{1}{x_n}\Big) = \sin(2\pi n) = 0, \qquad y_n = \frac{1}{2\pi n + \pi/2} \ \Rightarrow\ \sin\!\Big(\tfrac{1}{y_n}\Big) = 1. $$

$x_n \to 0$ 和 $y_n \to 0$ 都成立，但输出分别停在 $0$ 和 $1$。单一的极限值 $L$ 必须同时吸纳这两者，这不可能，所以极限不存在。与之对照的是 $\lim_{x\to 0} x\sin(\tfrac1x)$，它确实等于 $0$：在那里剧烈的振荡被乘以 $x \to 0$，第 3 节的夹逼定理（Squeeze Theorem）迫使其收敛。

The limit deliberately excludes the point $x = a$ so that it can measure the behavior of $f$ approaching $a$ independently of the value (or non-value) at $a$ itself. This separation is exactly what lets us later define the derivative as a limit of a difference quotient that is $0/0$ at the point of interest. If the limit were forced to use $f(a)$, calculus could not get off the ground.

极限刻意排除 $x = a$ 这一点，从而能够脱离 $a$ 处本身的取值（或无取值）来度量 $f$ 在趋近 $a$ 时的行为。正是这种分离，使我们日后能把导数（derivative）定义为差商的极限——而该差商在所关心的点恰好是 $0/0$。如果极限被迫使用 $f(a)$，微积分根本无从起步。

Evaluate $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$.

Substitution gives $0/0$. Multiply by the conjugate $\sqrt{x+4}+2$:

$$ \frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{(x+4) - 4}{x\big(\sqrt{x+4}+2\big)} = \frac{x}{x\big(\sqrt{x+4}+2\big)}. $$

Cancel the $x$ and substitute:

$$ \lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}. $$

求 $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$。

代入得到 $0/0$。乘以共轭式 $\sqrt{x+4}+2$：

$$ \frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{(x+4) - 4}{x\big(\sqrt{x+4}+2\big)} = \frac{x}{x\big(\sqrt{x+4}+2\big)}. $$

约去 $x$ 再代入：

$$ \lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}. $$

Evaluate $\lim_{x \to 0} \dfrac{\frac{1}{x+2} - \frac{1}{2}}{x}$.

Substitution gives $0/0$ again. The cure for a complex fraction is to combine the numerator over a common denominator first:

$$ \frac{1}{x+2} - \frac{1}{2} = \frac{2 - (x+2)}{2(x+2)} = \frac{-x}{2(x+2)}. $$

Now divide this by $x$, which is the same as multiplying by $\tfrac1x$:

$$ \frac{1}{x}\cdot\frac{-x}{2(x+2)} = \frac{-1}{2(x+2)}. $$

The troublesome $x$ has cancelled, so substitute:

$$ \lim_{x \to 0} \frac{-1}{2(x+2)} = \frac{-1}{2\cdot 2} = -\frac{1}{4}. $$

This is in fact the derivative of $1/x$ at $x = 2$ in disguise, which is why the same technique reappears throughout Unit A2.

求 $\lim_{x \to 0} \dfrac{\frac{1}{x+2} - \frac{1}{2}}{x}$。

代入又得到 $0/0$。处理繁分数的办法是先把分子通分合并：

$$ \frac{1}{x+2} - \frac{1}{2} = \frac{2 - (x+2)}{2(x+2)} = \frac{-x}{2(x+2)}. $$

再把它除以 $x$，也就是乘以 $\tfrac1x$：

$$ \frac{1}{x}\cdot\frac{-x}{2(x+2)} = \frac{-1}{2(x+2)}. $$

麻烦的 $x$ 已约去，于是代入：

$$ \lim_{x \to 0} \frac{-1}{2(x+2)} = \frac{-1}{2\cdot 2} = -\frac{1}{4}. $$

这其实是 $1/x$ 在 $x = 2$ 处导数（derivative）的伪装，这也是为什么同一技巧会贯穿第 A2 单元。

Evaluate $\lim_{x \to 3^+} \dfrac{x - 3}{\sqrt{x - 3}}$.

The expression is defined only for $x > 3$, since $\sqrt{x-3}$ needs a nonnegative argument, so a one-sided limit from the right is the only sensible question. For $x > 3$ write $x - 3 = \big(\sqrt{x-3}\big)^2$:

$$ \frac{x - 3}{\sqrt{x - 3}} = \frac{\big(\sqrt{x-3}\big)^2}{\sqrt{x-3}} = \sqrt{x - 3}. $$

Therefore $\lim_{x \to 3^+} \sqrt{x - 3} = 0$. The lesson is that the domain itself can dictate that only a one-sided limit exists; asking for the two-sided limit here would be meaningless because the function is not defined to the left of $3$.

求 $\lim_{x \to 3^+} \dfrac{x - 3}{\sqrt{x - 3}}$。

该表达式只在 $x > 3$ 时有定义，因为 $\sqrt{x-3}$ 要求被开方数非负，所以从右侧的单侧极限是唯一有意义的问题。对 $x > 3$，把 $x - 3 = \big(\sqrt{x-3}\big)^2$ 改写：

$$ \frac{x - 3}{\sqrt{x - 3}} = \frac{\big(\sqrt{x-3}\big)^2}{\sqrt{x-3}} = \sqrt{x - 3}. $$

于是 $\lim_{x \to 3^+} \sqrt{x - 3} = 0$。这里的教训是：定义域本身可以决定只存在单侧极限；在此处追问双侧极限毫无意义，因为函数在 $3$ 的左侧并无定义。

The limit laws are not axioms; they are theorems that follow from the epsilon-delta definition of Section 5. Here is the proof of the Sum Law, which is the template for the rest. Suppose $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M$. We claim $\lim_{x\to a}\big(f(x) + g(x)\big) = L + M$.

Let $\varepsilon > 0$. The target is $\big|(f(x)+g(x)) - (L+M)\big| < \varepsilon$. By the triangle inequality,

$$ \big|(f+g) - (L+M)\big| = \big|(f - L) + (g - M)\big| \le |f - L| + |g - M|. $$

The strategy is to make each piece smaller than $\varepsilon/2$. Since $\lim f = L$, there is a $\delta_1 > 0$ with $|f(x) - L| < \varepsilon/2$ whenever $0 < |x - a| < \delta_1$. Since $\lim g = M$, there is a $\delta_2 > 0$ with $|g(x) - M| < \varepsilon/2$ whenever $0 < |x - a| < \delta_2$. Take $\delta = \min\{\delta_1, \delta_2\}$. Then for $0 < |x - a| < \delta$ both bounds hold at once, so

$$ \big|(f+g) - (L+M)\big| \le |f - L| + |g - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$

Since $\varepsilon$ was arbitrary, the Sum Law holds. Splitting the tolerance in half so the pieces recombine to $\varepsilon$ is the recurring trick; you will see it again in the proof of the Product Law and in many later arguments.

极限法则不是公理；它们是从第 5 节的 epsilon-delta 定义推出的定理。下面是和法则（Sum Law）的证明，它是其余法则的模板。设 $\lim_{x\to a} f(x) = L$ 且 $\lim_{x\to a} g(x) = M$。我们断言 $\lim_{x\to a}\big(f(x) + g(x)\big) = L + M$。

设 $\varepsilon > 0$。目标是 $\big|(f(x)+g(x)) - (L+M)\big| < \varepsilon$。由三角不等式，

$$ \big|(f+g) - (L+M)\big| = \big|(f - L) + (g - M)\big| \le |f - L| + |g - M|. $$

策略是让每一项都小于 $\varepsilon/2$。由 $\lim f = L$，存在 $\delta_1 > 0$，使得只要 $0 < |x - a| < \delta_1$ 就有 $|f(x) - L| < \varepsilon/2$。由 $\lim g = M$，存在 $\delta_2 > 0$，使得只要 $0 < |x - a| < \delta_2$ 就有 $|g(x) - M| < \varepsilon/2$。取 $\delta = \min\{\delta_1, \delta_2\}$。则当 $0 < |x - a| < \delta$ 时两个界同时成立，故

$$ \big|(f+g) - (L+M)\big| \le |f - L| + |g - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$

由于 $\varepsilon$ 是任意的，和法则成立。把容差一分为二、使各项重新合成 $\varepsilon$ 是反复出现的技巧；你会在乘积法则（Product Rule 思路）的证明及后续许多论证中再次见到它。

For $0 < x < \pi/2$, compare areas inside the unit circle: the triangle with area $\tfrac12 \sin x$, the sector with area $\tfrac12 x$, and the larger triangle with area $\tfrac12 \tan x$. Then

$$ \sin x \le x \le \tan x. $$

Divide through by $\sin x > 0$ and invert (which flips the inequalities):

$$ 1 \ge \frac{\sin x}{x} \ge \cos x. $$

As $x \to 0^+$, $\cos x \to 1$, so by the Squeeze Theorem $\frac{\sin x}{x} \to 1$. Because $\frac{\sin x}{x}$ is even, the same holds as $x \to 0^-$, giving the two-sided limit $1$.

对 $0 < x < \pi/2$，比较单位圆内的三块面积：面积为 $\tfrac12 \sin x$ 的三角形、面积为 $\tfrac12 x$ 的扇形，以及面积为 $\tfrac12 \tan x$ 的较大三角形。于是

$$ \sin x \le x \le \tan x. $$

两边同除以 $\sin x > 0$ 再取倒数（不等号方向随之翻转）：

$$ 1 \ge \frac{\sin x}{x} \ge \cos x. $$

当 $x \to 0^+$ 时 $\cos x \to 1$，故由夹逼定理 $\frac{\sin x}{x} \to 1$。又因 $\frac{\sin x}{x}$ 是偶函数，当 $x \to 0^-$ 时同样成立，从而双侧极限为 $1$。

Show $\lim_{x \to 0} x^2 \sin\!\big(\tfrac{1}{x}\big) = 0$.

The factor $\sin(\tfrac1x)$ has no limit at $0$ (Worked Example 1.3), so the Product Law is useless here. Instead bound the bad factor. For every $x \neq 0$,

$$ -1 \le \sin\!\Big(\tfrac{1}{x}\Big) \le 1. $$

Multiply through by $x^2 \ge 0$, which preserves the inequalities:

$$ -x^2 \le x^2 \sin\!\Big(\tfrac{1}{x}\Big) \le x^2. $$

Both outer functions tend to $0$ as $x \to 0$, so the Squeeze Theorem forces the middle to $0$ as well. The oscillation never goes away, but the shrinking envelope $\pm x^2$ crushes it to zero.

证明 $\lim_{x \to 0} x^2 \sin\!\big(\tfrac{1}{x}\big) = 0$。

因子 $\sin(\tfrac1x)$ 在 $0$ 处没有极限（见例题 1.3），所以乘积法则在此无用。改为给这个捣乱的因子定界。对每个 $x \neq 0$，

$$ -1 \le \sin\!\Big(\tfrac{1}{x}\Big) \le 1. $$

两边同乘 $x^2 \ge 0$，不等号方向不变：

$$ -x^2 \le x^2 \sin\!\Big(\tfrac{1}{x}\Big) \le x^2. $$

当 $x \to 0$ 时外侧两个函数都趋于 $0$，于是夹逼定理迫使中间也趋于 $0$。振荡始终没有消失，但不断收缩的包络 $\pm x^2$ 把它压成了零。

Derive $\lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$ from the $\frac{\sin x}{x}$ limit.

Multiply numerator and denominator by the conjugate $1 + \cos x$ to convert the cosine into a sine via $1 - \cos^2 x = \sin^2 x$:

$$ \frac{1 - \cos x}{x}\cdot\frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x(1 + \cos x)} = \frac{\sin^2 x}{x(1 + \cos x)}. $$

Now regroup into a known piece times a piece that substitutes cleanly:

$$ \frac{\sin^2 x}{x(1 + \cos x)} = \frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}. $$

As $x \to 0$ the first factor tends to $1$ and the second tends to $\dfrac{0}{1 + 1} = 0$, so the product tends to $1 \cdot 0 = 0$. The same conjugate trick gives $\lim_{x\to 0}\dfrac{1 - \cos x}{x^2} = \tfrac12$, a result you will reuse when computing the second derivative of cosine.

从 $\frac{\sin x}{x}$ 极限推导 $\lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$。

分子分母同乘共轭式 $1 + \cos x$，借助 $1 - \cos^2 x = \sin^2 x$ 把余弦化成正弦：

$$ \frac{1 - \cos x}{x}\cdot\frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x(1 + \cos x)} = \frac{\sin^2 x}{x(1 + \cos x)}. $$

再重新分组成一个已知的部分乘以一个可以干净代入的部分：

$$ \frac{\sin^2 x}{x(1 + \cos x)} = \frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}. $$

当 $x \to 0$ 时第一个因子趋于 $1$，第二个趋于 $\dfrac{0}{1 + 1} = 0$，故乘积趋于 $1 \cdot 0 = 0$。同样的共轭技巧给出 $\lim_{x\to 0}\dfrac{1 - \cos x}{x^2} = \tfrac12$，这个结果在计算余弦的二阶导数时会再次用到。

Find $\lim_{x \to \infty} \dfrac{3x^2 - 5}{2x^2 + x}$. Divide numerator and denominator by the highest power $x^2$:

$$ \frac{3 - 5/x^2}{2 + 1/x} \xrightarrow{\ x \to \infty\ } \frac{3 - 0}{2 + 0} = \frac{3}{2}. $$

So $y = \tfrac32$ is a horizontal asymptote. Equal degrees give the ratio of leading coefficients.

求 $\lim_{x \to \infty} \dfrac{3x^2 - 5}{2x^2 + x}$。分子分母同除以最高次幂 $x^2$：

$$ \frac{3 - 5/x^2}{2 + 1/x} \xrightarrow{\ x \to \infty\ } \frac{3 - 0}{2 + 0} = \frac{3}{2}. $$

所以 $y = \tfrac32$ 是一条水平渐近线。次数相等时，极限等于首项系数之比。

Analyze $\lim_{x \to 2} \dfrac{1}{(x - 2)^2}$ and the two one-sided limits of $\lim_{x \to 2} \dfrac{1}{x - 2}$.

For the first, $(x-2)^2 > 0$ for all $x \neq 2$ and shrinks to $0$, so the reciprocal grows without bound from both sides:

$$ \lim_{x \to 2} \frac{1}{(x - 2)^2} = +\infty. $$

For the second, the sign of $x - 2$ matters. From the right, $x - 2 \to 0^+$, so the reciprocal tends to $+\infty$; from the left, $x - 2 \to 0^-$, so it tends to $-\infty$:

$$ \lim_{x \to 2^+} \frac{1}{x - 2} = +\infty, \qquad \lim_{x \to 2^-} \frac{1}{x - 2} = -\infty. $$

Because the two sides disagree even in their infinite behavior, the two-sided limit does not exist (not even as $+\infty$). The line $x = 2$ is a vertical asymptote in both cases; the difference is whether the function dives the same way or opposite ways on the two sides.

分析 $\lim_{x \to 2} \dfrac{1}{(x - 2)^2}$，以及 $\lim_{x \to 2} \dfrac{1}{x - 2}$ 的两个单侧极限。

对第一个，$(x-2)^2 > 0$ 对所有 $x \neq 2$ 成立且趋于 $0$，故其倒数从两侧都无界增大：

$$ \lim_{x \to 2} \frac{1}{(x - 2)^2} = +\infty. $$

对第二个，$x - 2$ 的符号很关键。从右侧 $x - 2 \to 0^+$，故倒数趋于 $+\infty$；从左侧 $x - 2 \to 0^-$，故趋于 $-\infty$：

$$ \lim_{x \to 2^+} \frac{1}{x - 2} = +\infty, \qquad \lim_{x \to 2^-} \frac{1}{x - 2} = -\infty. $$

由于两侧连无穷行为都不一致，双侧极限不存在（甚至不能记为 $+\infty$）。两种情形里直线 $x = 2$ 都是垂直渐近线；区别在于函数在两侧是朝同一方向还是相反方向发散。

Find $\lim_{x \to \infty} \dfrac{\sqrt{9x^2 + 1}}{x + 5}$.

The highest power inside the root is $x^2$, whose square root is $|x|$. For $x \to \infty$ we have $x > 0$, so $|x| = x$. Factor $x^2$ out of the radical:

$$ \sqrt{9x^2 + 1} = \sqrt{x^2\Big(9 + \tfrac{1}{x^2}\Big)} = |x|\sqrt{9 + \tfrac{1}{x^2}} = x\sqrt{9 + \tfrac{1}{x^2}} \quad (x > 0). $$

Divide numerator and denominator by $x$:

$$ \frac{x\sqrt{9 + 1/x^2}}{x(1 + 5/x)} = \frac{\sqrt{9 + 1/x^2}}{1 + 5/x} \xrightarrow{\ x\to\infty\ } \frac{\sqrt{9}}{1} = 3. $$

The horizontal asymptote is $y = 3$. Had the question asked for $x \to -\infty$, the step $|x| = x$ would have become $|x| = -x$, and the limit would have been $-3$; sign care with $\sqrt{x^2} = |x|$ is the whole subtlety here.

求 $\lim_{x \to \infty} \dfrac{\sqrt{9x^2 + 1}}{x + 5}$。

根号内最高次幂是 $x^2$，其平方根是 $|x|$。当 $x \to \infty$ 时 $x > 0$，故 $|x| = x$。从根式中提出 $x^2$：

$$ \sqrt{9x^2 + 1} = \sqrt{x^2\Big(9 + \tfrac{1}{x^2}\Big)} = |x|\sqrt{9 + \tfrac{1}{x^2}} = x\sqrt{9 + \tfrac{1}{x^2}} \quad (x > 0). $$

分子分母同除以 $x$：

$$ \frac{x\sqrt{9 + 1/x^2}}{x(1 + 5/x)} = \frac{\sqrt{9 + 1/x^2}}{1 + 5/x} \xrightarrow{\ x\to\infty\ } \frac{\sqrt{9}}{1} = 3. $$

水平渐近线是 $y = 3$。若问的是 $x \to -\infty$，则 $|x| = x$ 这一步会变成 $|x| = -x$，极限将是 $-3$；对 $\sqrt{x^2} = |x|$ 的符号谨慎正是这里的全部微妙之处。

Prove $\lim_{x \to 4} (3x - 5) = 7$.

Scratch work. We need $|(3x - 5) - 7| < \varepsilon$, i.e. $|3x - 12| = 3|x - 4| < \varepsilon$, i.e. $|x - 4| < \varepsilon/3$. That tells us the right choice is $\delta = \varepsilon/3$.

Proof. Let $\varepsilon > 0$ and set $\delta = \varepsilon/3$. If $0 < |x - 4| < \delta$, then

$$ |(3x - 5) - 7| = 3|x - 4| < 3\delta = 3\cdot\frac{\varepsilon}{3} = \varepsilon. $$

Since $\varepsilon$ was arbitrary, the limit is $7$.

证明 $\lim_{x \to 4} (3x - 5) = 7$。

草稿。我们需要 $|(3x - 5) - 7| < \varepsilon$，即 $|3x - 12| = 3|x - 4| < \varepsilon$，即 $|x - 4| < \varepsilon/3$。这告诉我们正确的选择是 $\delta = \varepsilon/3$。

证明。设 $\varepsilon > 0$，令 $\delta = \varepsilon/3$。若 $0 < |x - 4| < \delta$，则

$$ |(3x - 5) - 7| = 3|x - 4| < 3\delta = 3\cdot\frac{\varepsilon}{3} = \varepsilon. $$

由于 $\varepsilon$ 是任意的，极限为 $7$。

Prove $\lim_{x \to 2} x^2 = 4$. We need $|x^2 - 4| = |x - 2|\,|x + 2| < \varepsilon$. The factor $|x+2|$ is not constant, so first restrict $\delta \le 1$. Then $|x - 2| < 1$ gives $1 < x < 3$, hence $|x + 2| < 5$. Now

$$ |x^2 - 4| = |x-2|\,|x+2| < 5\,|x - 2|. $$

To force this below $\varepsilon$, also require $|x - 2| < \varepsilon/5$. Choose $\delta = \min\{1,\ \varepsilon/5\}$. Then both bounds hold and $|x^2 - 4| < 5\cdot(\varepsilon/5) = \varepsilon$. The $\min$ is the standard device whenever the slack factor depends on $x$.

证明 $\lim_{x \to 2} x^2 = 4$。我们需要 $|x^2 - 4| = |x - 2|\,|x + 2| < \varepsilon$。因子 $|x+2|$ 不是常数，所以先限制 $\delta \le 1$。则 $|x - 2| < 1$ 给出 $1 < x < 3$，故 $|x + 2| < 5$。于是

$$ |x^2 - 4| = |x-2|\,|x+2| < 5\,|x - 2|. $$

为把它压到 $\varepsilon$ 以下，再要求 $|x - 2| < \varepsilon/5$。取 $\delta = \min\{1,\ \varepsilon/5\}$。则两个界同时成立，且 $|x^2 - 4| < 5\cdot(\varepsilon/5) = \varepsilon$。每当那个伸缩因子依赖于 $x$ 时，取 $\min$ 是标准手段。

Prove $\lim_{x \to 3} \dfrac{1}{x} = \dfrac{1}{3}$.

Scratch work. We need to control

$$ \left|\frac{1}{x} - \frac{1}{3}\right| = \left|\frac{3 - x}{3x}\right| = \frac{|x - 3|}{3\,|x|}. $$

The factor $\tfrac{1}{|x|}$ blows up if $x$ is allowed near $0$, so first keep $x$ away from $0$. Require $\delta \le \tfrac32$, so $|x - 3| < \tfrac32$ gives $\tfrac32 < x < \tfrac92$, hence $|x| > \tfrac32$ and $\tfrac{1}{|x|} < \tfrac23$. Then

$$ \frac{|x - 3|}{3\,|x|} < \frac{|x - 3|}{3}\cdot\frac{2}{3} = \frac{2}{9}\,|x - 3|. $$

To push this below $\varepsilon$, also require $|x - 3| < \tfrac{9}{2}\varepsilon$.

Proof. Let $\varepsilon > 0$ and set $\delta = \min\big\{\tfrac32,\ \tfrac92\varepsilon\big\}$. If $0 < |x - 3| < \delta$, then $|x| > \tfrac32$ from the first bound, and

$$ \left|\frac{1}{x} - \frac{1}{3}\right| = \frac{|x - 3|}{3\,|x|} < \frac{2}{9}\,|x - 3| < \frac{2}{9}\cdot\frac{9}{2}\varepsilon = \varepsilon. $$

Since $\varepsilon$ was arbitrary, the limit is $\tfrac13$. The new idea beyond the quadratic case is that here $\delta$ does double duty: it bounds the slack factor and keeps $x$ inside the domain where $\tfrac1x$ is well behaved.

证明 $\lim_{x \to 3} \dfrac{1}{x} = \dfrac{1}{3}$。

草稿。我们需要控制

$$ \left|\frac{1}{x} - \frac{1}{3}\right| = \left|\frac{3 - x}{3x}\right| = \frac{|x - 3|}{3\,|x|}. $$

若允许 $x$ 靠近 $0$，因子 $\tfrac{1}{|x|}$ 会发散，所以先让 $x$ 远离 $0$。要求 $\delta \le \tfrac32$，则 $|x - 3| < \tfrac32$ 给出 $\tfrac32 < x < \tfrac92$，故 $|x| > \tfrac32$ 且 $\tfrac{1}{|x|} < \tfrac23$。于是

$$ \frac{|x - 3|}{3\,|x|} < \frac{|x - 3|}{3}\cdot\frac{2}{3} = \frac{2}{9}\,|x - 3|. $$

为把它压到 $\varepsilon$ 以下，再要求 $|x - 3| < \tfrac{9}{2}\varepsilon$。

证明。设 $\varepsilon > 0$，令 $\delta = \min\big\{\tfrac32,\ \tfrac92\varepsilon\big\}$。若 $0 < |x - 3| < \delta$，则由第一个界 $|x| > \tfrac32$，且

$$ \left|\frac{1}{x} - \frac{1}{3}\right| = \frac{|x - 3|}{3\,|x|} < \frac{2}{9}\,|x - 3| < \frac{2}{9}\cdot\frac{9}{2}\varepsilon = \varepsilon. $$

由于 $\varepsilon$ 是任意的，极限为 $\tfrac13$。相较二次情形，这里的新意是 $\delta$ 身兼两职：既给伸缩因子定界，又把 $x$ 保持在 $\tfrac1x$ 表现良好的定义域之内。

To show a limit does not equal a proposed value, negate the definition. The statement $\lim_{x\to a} f(x) = L$ fails when

$$ \exists\, \varepsilon > 0 \ \text{ such that } \ \forall\, \delta > 0,\ \exists\, x \text{ with } 0 < |x - a| < \delta \text{ and } |f(x) - L| \ge \varepsilon. $$

In words: there is one stubborn tolerance $\varepsilon$ that no $\delta$ can satisfy, because inside every punctured neighborhood of $a$ some point is thrown at least $\varepsilon$ away from $L$.

Apply this to disprove $\lim_{x\to 0}\dfrac{|x|}{x} = 1$. Take $\varepsilon = 1$. For any $\delta > 0$, pick a point on the left, say $x = -\tfrac{\delta}{2}$, which satisfies $0 < |x| < \delta$. There $f(x) = -1$, so

$$ |f(x) - 1| = |-1 - 1| = 2 \ge 1 = \varepsilon. $$

Since this works for every $\delta$, the limit is not $1$. The same argument rules out every candidate $L$, which is the rigorous version of "the one-sided limits disagree, so no limit exists."

要证明一个极限不等于某个提议值，就否定定义。当满足下式时，断言 $\lim_{x\to a} f(x) = L$ 失败：

$$ \exists\, \varepsilon > 0 \ \text{ such that } \ \forall\, \delta > 0,\ \exists\, x \text{ with } 0 < |x - a| < \delta \text{ and } |f(x) - L| \ge \varepsilon. $$

用文字说：存在一个顽固的容差 $\varepsilon$ 是任何 $\delta$ 都满足不了的，因为在 $a$ 的每个去心邻域内，总有某个点被抛到离 $L$ 至少 $\varepsilon$ 远处。

用它来反证 $\lim_{x\to 0}\dfrac{|x|}{x} = 1$。取 $\varepsilon = 1$。对任意 $\delta > 0$，在左侧取一点，比如 $x = -\tfrac{\delta}{2}$，它满足 $0 < |x| < \delta$。在那里 $f(x) = -1$，故

$$ |f(x) - 1| = |-1 - 1| = 2 \ge 1 = \varepsilon. $$

由于这对每个 $\delta$ 都成立，极限不是 $1$。同样的论证排除了每个候选 $L$，这正是 "两个单侧极限不一致，故极限不存在" 的严格版本。

Show $x^3 + x - 1 = 0$ has a solution in $(0, 1)$.

Let $f(x) = x^3 + x - 1$, a polynomial, hence continuous on $[0,1]$. Then $f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $0$ lies between $f(0)$ and $f(1)$, the IVT guarantees a $c \in (0,1)$ with $f(c) = 0$.

The theorem is an existence statement: it proves a root is there without locating it. Bisection then narrows the interval as far as desired.

证明 $x^3 + x - 1 = 0$ 在 $(0, 1)$ 内有解。

设 $f(x) = x^3 + x - 1$，它是多项式，故在 $[0,1]$ 上连续。则 $f(0) = -1 < 0$ 且 $f(1) = 1 > 0$。由于 $0$ 介于 $f(0)$ 与 $f(1)$ 之间，介值定理（IVT）保证存在 $c \in (0,1)$ 使 $f(c) = 0$。

该定理是一个存在性陈述：它证明根在那里，却不定位它。随后用二分法可把区间收窄到任意精度。

Find the value of $k$ that makes

$$ f(x) = \begin{cases} \dfrac{x^2 - 9}{x - 3}, & x \neq 3 \\[6pt] k, & x = 3 \end{cases} $$

continuous at $x = 3$.

Continuity requires $\lim_{x\to 3} f(x) = f(3) = k$. The limit was computed in Worked Example 1.1: cancelling the factor $x - 3$ gives $\lim_{x\to 3}(x + 3) = 6$. So we must set

$$ k = 6. $$

With $k = 6$ the removable hole is filled and $f$ becomes continuous everywhere. This is the precise meaning of "removable": there exists a single value we could assign at the point to repair the break. If the limit had failed to exist, as in a jump, no choice of $k$ could make $f$ continuous there.

求使

$$ f(x) = \begin{cases} \dfrac{x^2 - 9}{x - 3}, & x \neq 3 \\[6pt] k, & x = 3 \end{cases} $$

在 $x = 3$ 处连续的 $k$ 值。

连续要求 $\lim_{x\to 3} f(x) = f(3) = k$。该极限已在例题 1.1 中算出：约去因子 $x - 3$ 得 $\lim_{x\to 3}(x + 3) = 6$。所以必须令

$$ k = 6. $$

当 $k = 6$ 时，可去的空洞被填补，$f$ 处处连续。这正是 "可去" 的精确含义：存在单一的值，我们可以在该点赋上以修复断裂。如果极限本就不存在（例如跳跃），任何 $k$ 的取值都无法使 $f$ 在那里连续。

Show that $\cos x = x$ has a solution in $\big(0, \tfrac{\pi}{2}\big)$.

Direct algebra cannot solve this transcendental equation, so reframe it as a root problem. Let $g(x) = \cos x - x$, which is continuous on $\big[0, \tfrac{\pi}{2}\big]$ as a difference of continuous functions. Evaluate the endpoints:

$$ g(0) = \cos 0 - 0 = 1 > 0, \qquad g\!\Big(\tfrac{\pi}{2}\Big) = \cos\tfrac{\pi}{2} - \tfrac{\pi}{2} = -\tfrac{\pi}{2} < 0. $$

Since $0$ lies between $g(0)$ and $g(\tfrac{\pi}{2})$, the IVT gives a $c \in \big(0, \tfrac{\pi}{2}\big)$ with $g(c) = 0$, that is $\cos c = c$. This is the standard proof that the equation $\cos x = x$ has a fixed point; numerically $c \approx 0.739$, but the IVT delivers existence with no computation at all.

证明 $\cos x = x$ 在 $\big(0, \tfrac{\pi}{2}\big)$ 内有解。

直接代数无法解这个超越方程，所以把它改写成求根问题。设 $g(x) = \cos x - x$，作为两个连续函数之差，它在 $\big[0, \tfrac{\pi}{2}\big]$ 上连续。计算端点：

$$ g(0) = \cos 0 - 0 = 1 > 0, \qquad g\!\Big(\tfrac{\pi}{2}\Big) = \cos\tfrac{\pi}{2} - \tfrac{\pi}{2} = -\tfrac{\pi}{2} < 0. $$

由于 $0$ 介于 $g(0)$ 与 $g(\tfrac{\pi}{2})$ 之间，介值定理给出 $c \in \big(0, \tfrac{\pi}{2}\big)$ 使 $g(c) = 0$，即 $\cos c = c$。这是方程 $\cos x = x$ 有不动点的标准证明；数值上 $c \approx 0.739$，但介值定理无需任何计算就给出了存在性。

Each hypothesis of the EVT is doing real work, which the following counterexamples make concrete.

Dropping closedness. On $(0, 1]$ the function $f(x) = 1/x$ is continuous, but as $x \to 0^+$ it grows without bound, so it has no maximum. The supremum is $+\infty$ and is never attained. The closed left endpoint is exactly what the EVT needs and the open interval denies.

Dropping boundedness. On the unbounded interval $[0, \infty)$ the continuous function $f(x) = x$ has no maximum. A finite interval is essential.

Dropping continuity. On the closed interval $[0, 1]$ define $f(x) = x$ for $x < 1$ and $f(1) = 0$. The values approach $1$ but never reach it, since the only point that could supply the value $1$ has been redefined to $0$. The supremum $1$ is not attained, so no maximum exists. A single discontinuity is enough to break the conclusion.

The three counterexamples show that continuity, closedness, and boundedness are jointly, not redundantly, responsible for the guarantee.

极值定理的每个前提都在实打实地起作用，下面的反例把这一点具体化。

去掉闭性。在 $(0, 1]$ 上，函数 $f(x) = 1/x$ 连续，但当 $x \to 0^+$ 时它无界增大，故没有最大值。上确界是 $+\infty$ 且永不取到。闭的左端点正是极值定理所需、而开区间所欠缺的。

去掉有界性。在无界区间 $[0, \infty)$ 上，连续函数 $f(x) = x$ 没有最大值。有限区间是本质性的。

去掉连续性。在闭区间 $[0, 1]$ 上，定义当 $x < 1$ 时 $f(x) = x$，且 $f(1) = 0$。函数值趋近 $1$ 却从不到达，因为唯一能提供值 $1$ 的那个点已被重新定义为 $0$。上确界 $1$ 取不到，故没有最大值。单一的间断就足以破坏结论。

这三个反例表明：连续性、闭性与有界性是共同地、而非冗余地承担起这一保证的。

The EVT has two halves: a continuous function on $[a,b]$ is bounded, and it actually attains its bounds. The boundedness half already reveals why the closed interval is indispensable.

Suppose, for contradiction, that $f$ is continuous on $[a,b]$ but unbounded above. Then for each $n$ there is a point $x_n \in [a,b]$ with $f(x_n) > n$. The sequence $(x_n)$ lives in the closed bounded interval $[a,b]$, so by the Bolzano-Weierstrass property it has a subsequence $x_{n_k}$ converging to some limit $c \in [a,b]$. Here the closedness matters: the limit $c$ cannot escape the interval. By continuity, $f(x_{n_k}) \to f(c)$, a finite number. But by construction $f(x_{n_k}) > n_k \to \infty$, so $f(x_{n_k})$ is unbounded. A sequence cannot converge to a finite value and diverge to infinity at once, a contradiction. Hence $f$ is bounded above; the same argument bounds it below.

Once $f$ is known to be bounded, let $M = \sup_{[a,b]} f$. A second argument, again using a convergent subsequence and continuity, produces a point where $f$ equals $M$, which completes the EVT. The point to retain is that the closed bounded interval is precisely what keeps the limit point $c$ inside the domain, which is why both hypotheses, dropped in Worked Example 7.1, were fatal.

极值定理有两半：$[a,b]$ 上的连续函数是有界的，并且它确实取到了它的界。有界这一半已经揭示了为何闭区间不可或缺。

反证：设 $f$ 在 $[a,b]$ 上连续但向上无界。则对每个 $n$ 存在一点 $x_n \in [a,b]$ 使 $f(x_n) > n$。数列 $(x_n)$ 落在有界闭区间 $[a,b]$ 内，故由 Bolzano-Weierstrass 性质，它有一个收敛到某极限 $c \in [a,b]$ 的子列 $x_{n_k}$。这里闭性很关键：极限 $c$ 无法逃出区间。由连续性，$f(x_{n_k}) \to f(c)$，一个有限数。但由构造 $f(x_{n_k}) > n_k \to \infty$，故 $f(x_{n_k})$ 无界。一个数列不可能同时收敛到有限值又发散（divergence）到无穷，矛盾。于是 $f$ 向上有界；同样的论证给出向下有界。

一旦知道 $f$ 有界，令 $M = \sup_{[a,b]} f$。再用一次收敛子列与连续性的论证，可得到一个使 $f$ 等于 $M$ 的点，从而完成极值定理。要记住的要点是：有界闭区间正是把极限点 $c$ 留在定义域内的东西，这也是为何在例题 7.1 中去掉任一前提都是致命的。

The whole machinery of this unit exists to support one definition. The derivative is

$$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}, $$

a limit whose difference quotient is exactly $0/0$ at $h = 0$. Everything you practiced here (recognizing indeterminate forms, simplifying before substituting, and trusting the limit even where the expression is undefined at the point) is precisely the skill set Unit A2 requires. Limits are not a preliminary to calculus; they are its definition.

本单元的整套机器都是为支撑一个定义而存在的。导数（derivative）是

$$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}, $$

一个其差商在 $h = 0$ 处恰为 $0/0$ 的极限。你在这里练就的一切（辨认不定式、代入前先化简、即使表达式在该点无定义也信任极限）正是第 A2 单元所需的技能组合。极限不是微积分的前奏；它就是微积分的定义。