Unit A6: Linear Approximation, Differentials, and L’Hopital第 A6 单元：线性近似、微分与洛必达

Let $f(x) = \sqrt{x}$ and pick the convenient base point $a = 4$, where the square root is exact. Then $f(4) = 2$ and $f'(x) = \tfrac{1}{2\sqrt{x}}$, so $f'(4) = \tfrac{1}{4}$.

$$ L(x) = 2 + \tfrac{1}{4}(x - 4). $$

Evaluating at $x = 4.1$,

$$ \sqrt{4.1} \approx L(4.1) = 2 + \tfrac{1}{4}(0.1) = 2.025. $$

The true value is $2.024845\ldots$, so the linear estimate is correct to four decimal places. The error is positive because $\sqrt{x}$ is concave down, so the tangent line lies above the curve.

取 $f(x) = \sqrt{x}$，并选取便于计算的基点 $a = 4$，那里平方根是精确的。于是 $f(4) = 2$，$f'(x) = \tfrac{1}{2\sqrt{x}}$，所以 $f'(4) = \tfrac{1}{4}$。

$$ L(x) = 2 + \tfrac{1}{4}(x - 4). $$

在 $x = 4.1$ 处求值，

$$ \sqrt{4.1} \approx L(4.1) = 2 + \tfrac{1}{4}(0.1) = 2.025. $$

真值是 $2.024845\ldots$，所以这个线性估计精确到四位小数。误差为正，因为 $\sqrt{x}$ 是凹（向下凹）的，所以切线位于曲线上方。

Write $(1.02)^{10} = (1 + x)^{10}$ with $x = 0.02$, and use $(1 + x)^k \approx 1 + kx$ with $k = 10$.

$$ (1.02)^{10} \approx 1 + 10(0.02) = 1.20. $$

The true value is $1.2190\ldots$. The estimate $1.20$ is low because the curve is convex; the error of about $0.019$ is governed by the quadratic term $\binom{10}{2}x^2 = 45(0.0004) = 0.018$, which the linear model discards.

把 $(1.02)^{10} = (1 + x)^{10}$ 写成 $x = 0.02$ 的形式，并用 $(1 + x)^k \approx 1 + kx$（取 $k = 10$）。

$$ (1.02)^{10} \approx 1 + 10(0.02) = 1.20. $$

真值是 $1.2190\ldots$。估计值 $1.20$ 偏低，因为曲线是凸的；约 $0.019$ 的误差主要由二次项 $\binom{10}{2}x^2 = 45(0.0004) = 0.018$ 决定，而线性模型把它丢掉了。

First the easy base point. With $f(x) = \sin x$ at $a = 0$ we have $f(0) = 0$ and $f'(0) = \cos 0 = 1$, so $L(x) = x$ and

$$ \sin(0.1) \approx 0.1, \qquad \text{true value } 0.0998334\ldots $$

The error is only about $1.7 \times 10^{-4}$, again governed by the discarded cubic term $-x^3/6 = -1.67\times 10^{-4}$.

Now a case where the right base point matters. To estimate $\cos(31^\circ)$, work in radians and choose $a = \pi/6 = 30^\circ$, where the cosine is known exactly. Here $f(x) = \cos x$, $f(\pi/6) = \tfrac{\sqrt{3}}{2}$, and $f'(x) = -\sin x$, so $f'(\pi/6) = -\tfrac12$. The increment is $x - a = 1^\circ = \pi/180 \approx 0.01745$ rad:

$$ \cos(31^\circ) \approx \tfrac{\sqrt3}{2} - \tfrac12(0.01745) = 0.86603 - 0.00873 = 0.85730. $$

The true value is $0.85717\ldots$, accurate to four decimals. Notice the radian conversion is essential: $f'(x) = -\sin x$ is the derivative only when $x$ is measured in radians.

先看简单的基点。取 $f(x) = \sin x$，在 $a = 0$ 处有 $f(0) = 0$、$f'(0) = \cos 0 = 1$，所以 $L(x) = x$，于是

$$ \sin(0.1) \approx 0.1, \qquad \text{真值 } 0.0998334\ldots $$

误差仅约 $1.7 \times 10^{-4}$，同样由被丢弃的三次项 $-x^3/6 = -1.67\times 10^{-4}$ 决定。

再看一个选对基点很关键的情形。要估计 $\cos(31^\circ)$，需用弧度计算，并选 $a = \pi/6 = 30^\circ$，那里余弦值已知且精确。此时 $f(x) = \cos x$，$f(\pi/6) = \tfrac{\sqrt{3}}{2}$，$f'(x) = -\sin x$，所以 $f'(\pi/6) = -\tfrac12$。增量为 $x - a = 1^\circ = \pi/180 \approx 0.01745$ 弧度：

$$ \cos(31^\circ) \approx \tfrac{\sqrt3}{2} - \tfrac12(0.01745) = 0.86603 - 0.00873 = 0.85730. $$

真值是 $0.85717\ldots$，精确到四位小数。注意弧度换算是必不可少的：$f'(x) = -\sin x$ 只有在 $x$ 以弧度度量时才是导数。

We prove the theorem stated above: among all linear functions $\ell(x) = c_0 + c_1(x-a)$, the linearization $L(x) = f(a) + f'(a)(x-a)$ is the unique one whose error is $o(x-a)$ as $x \to a$.

Step 1: $L$ works. Write the error $E(x) = f(x) - L(x) = f(x) - f(a) - f'(a)(x-a)$. For $x \neq a$,

$$ \frac{E(x)}{x - a} = \frac{f(x) - f(a)}{x - a} - f'(a) \ \longrightarrow\ f'(a) - f'(a) = 0 $$

by the definition of the derivative. Hence $E(x) = o(x-a)$.

Step 2: uniqueness. Suppose some $\ell(x) = c_0 + c_1(x-a)$ also has $f(x) - \ell(x) = o(x-a)$. Subtracting the two error statements, $L(x) - \ell(x) = o(x-a)$. But $L(x) - \ell(x) = (f(a) - c_0) + (f'(a) - c_1)(x-a)$ is itself linear. Letting $x \to a$ in $L(x) - \ell(x) \to 0$ forces $c_0 = f(a)$. With that constant gone, $\frac{(f'(a)-c_1)(x-a)}{x-a} = f'(a) - c_1 \to 0$ forces $c_1 = f'(a)$. Thus $\ell = L$. The simultaneous matching of value and slope is exactly what makes the tangent line special.

我们来证明上面叙述的定理：在所有线性函数 $\ell(x) = c_0 + c_1(x-a)$ 中，线性化 $L(x) = f(a) + f'(a)(x-a)$ 是唯一一个当 $x \to a$ 时误差为 $o(x-a)$ 的。

第 1 步：$L$ 成立。 记误差 $E(x) = f(x) - L(x) = f(x) - f(a) - f'(a)(x-a)$。对 $x \neq a$，

$$ \frac{E(x)}{x - a} = \frac{f(x) - f(a)}{x - a} - f'(a) \ \longrightarrow\ f'(a) - f'(a) = 0 $$

这是由导数的定义得到的。因此 $E(x) = o(x-a)$。

第 2 步：唯一性。 假设某个 $\ell(x) = c_0 + c_1(x-a)$ 也满足 $f(x) - \ell(x) = o(x-a)$。把两个误差式相减，得 $L(x) - \ell(x) = o(x-a)$。但 $L(x) - \ell(x) = (f(a) - c_0) + (f'(a) - c_1)(x-a)$ 本身就是线性的。在 $L(x) - \ell(x) \to 0$ 中令 $x \to a$，迫使 $c_0 = f(a)$。这个常数消去后，$\frac{(f'(a)-c_1)(x-a)}{x-a} = f'(a) - c_1 \to 0$ 迫使 $c_1 = f'(a)$。于是 $\ell = L$。同时匹配值和斜率，正是切线特殊之处。

A circular plate has radius measured as $r = 12$ cm with a possible error of $dr = 0.05$ cm. Estimate the resulting error in the computed area $A = \pi r^2$.

$$ dA = \frac{dA}{dr}\,dr = 2\pi r\,dr = 2\pi (12)(0.05) = 1.2\pi \approx 3.77 \text{ cm}^2. $$

The relative error is

$$ \frac{dA}{A} = \frac{2\pi r\,dr}{\pi r^2} = \frac{2\,dr}{r} = \frac{2(0.05)}{12} \approx 0.0083, $$

so about $0.83\%$. Note the relative area error is twice the relative radius error, a direct consequence of the exponent $2$ in $A = \pi r^2$.

一块圆盘测得半径 $r = 12$ cm，可能误差为 $dr = 0.05$ cm。估计由此导致的面积 $A = \pi r^2$ 的误差。

$$ dA = \frac{dA}{dr}\,dr = 2\pi r\,dr = 2\pi (12)(0.05) = 1.2\pi \approx 3.77 \text{ cm}^2. $$

相对误差为

$$ \frac{dA}{A} = \frac{2\pi r\,dr}{\pi r^2} = \frac{2\,dr}{r} = \frac{2(0.05)}{12} \approx 0.0083, $$

即约 $0.83\%$。注意面积的相对误差是半径相对误差的两倍，这是 $A = \pi r^2$ 中指数 $2$ 的直接结果。

Differentiability at $x$ means the limit defining $f'(x)$ exists. Define the error function

$$ \varepsilon(dx) = \frac{f(x + dx) - f(x)}{dx} - f'(x) \quad (dx \neq 0), \qquad \varepsilon(0) = 0. $$

By the definition of the derivative, $\varepsilon(dx) \to 0$ as $dx \to 0$. Multiplying through by $dx$,

$$ \Delta y = f(x + dx) - f(x) = f'(x)\,dx + \varepsilon(dx)\,dx = dy + \varepsilon(dx)\,dx. $$

Since $\varepsilon(dx) \to 0$, the remainder $\varepsilon(dx)\,dx$ is $o(dx)$: it shrinks faster than $dx$ itself. This is precisely the statement that the tangent line is the first-order model of $f$, and it is the foundation for every estimate in this unit.

$f$ 在 $x$ 处可微意味着定义 $f'(x)$ 的极限（limit）存在。定义误差函数

$$ \varepsilon(dx) = \frac{f(x + dx) - f(x)}{dx} - f'(x) \quad (dx \neq 0), \qquad \varepsilon(0) = 0. $$

由导数的定义，当 $dx \to 0$ 时 $\varepsilon(dx) \to 0$。两边乘以 $dx$，

$$ \Delta y = f(x + dx) - f(x) = f'(x)\,dx + \varepsilon(dx)\,dx = dy + \varepsilon(dx)\,dx. $$

由于 $\varepsilon(dx) \to 0$，余项 $\varepsilon(dx)\,dx$ 是 $o(dx)$：它比 $dx$ 本身更快地缩小。这正是“切线是 $f$ 的一阶模型”这一说法，也是本单元每一个估计的基础。

The period of a simple pendulum is $T = 2\pi\sqrt{L/g}$. If the length $L$ is known to within $1.5\%$, what is the resulting uncertainty in $T$? Write $T = 2\pi g^{-1/2} L^{1/2}$ and take the differential:

$$ dT = 2\pi g^{-1/2} \cdot \tfrac12 L^{-1/2}\,dL = \frac{1}{2}\,\frac{2\pi\sqrt{L/g}}{L}\,dL = \frac{T}{2}\cdot\frac{dL}{L}. $$

Dividing by $T$ gives the clean relative relation

$$ \frac{dT}{T} = \frac{1}{2}\,\frac{dL}{L} = \frac12 (0.015) = 0.0075, $$

so the period is uncertain by about $0.75\%$, half the relative error in the length. The general rule for a power law $y = k x^p$ is $\dfrac{dy}{y} = p\,\dfrac{dx}{x}$: relative errors scale by the exponent, which is why the disk area in Example 2.1 doubled the radius error and this pendulum halves the length error.

单摆的周期为 $T = 2\pi\sqrt{L/g}$。若长度 $L$ 的已知精度在 $1.5\%$ 以内，由此导致 $T$ 的不确定度是多少？把 $T = 2\pi g^{-1/2} L^{1/2}$ 写出并取微分：

$$ dT = 2\pi g^{-1/2} \cdot \tfrac12 L^{-1/2}\,dL = \frac{1}{2}\,\frac{2\pi\sqrt{L/g}}{L}\,dL = \frac{T}{2}\cdot\frac{dL}{L}. $$

除以 $T$ 得到简洁的相对关系

$$ \frac{dT}{T} = \frac{1}{2}\,\frac{dL}{L} = \frac12 (0.015) = 0.0075, $$

所以周期的不确定度约为 $0.75\%$，是长度相对误差的一半。对幂律 $y = k x^p$ 的一般规则是 $\dfrac{dy}{y} = p\,\dfrac{dx}{x}$：相对误差按指数缩放，这就是为什么例题 2.1 中圆盘面积把半径误差翻倍，而这个单摆把长度误差减半。

Let $y = f(x) = x^3$ at $x = 2$ with $dx = \Delta x = 0.1$. Compare the differential with the true change.

$$ dy = 3x^2\,dx = 3(4)(0.1) = 1.2, $$ $$ \Delta y = f(2.1) - f(2) = 9.261 - 8 = 1.261. $$

The differential captures $1.2$ of the actual $1.261$; the leftover $0.061$ is the higher-order part. Indeed the exact expansion is $\Delta y = 3x^2\,dx + 3x\,(dx)^2 + (dx)^3 = 1.2 + 0.06 + 0.001 = 1.261$, and the discarded terms $3x(dx)^2 + (dx)^3 = 0.061$ are precisely the $o(dx)$ remainder from Section 1. As $dx$ shrinks, that remainder vanishes faster than $dy$ itself.

取 $y = f(x) = x^3$，在 $x = 2$ 处，$dx = \Delta x = 0.1$。比较微分与真实变化。

$$ dy = 3x^2\,dx = 3(4)(0.1) = 1.2, $$ $$ \Delta y = f(2.1) - f(2) = 9.261 - 8 = 1.261. $$

微分捕捉了实际值 $1.261$ 中的 $1.2$；余下的 $0.061$ 是高阶部分。事实上精确展开为 $\Delta y = 3x^2\,dx + 3x\,(dx)^2 + (dx)^3 = 1.2 + 0.06 + 0.001 = 1.261$，而被丢弃的项 $3x(dx)^2 + (dx)^3 = 0.061$ 恰好是第 1 节中的 $o(dx)$ 余项。当 $dx$ 缩小时，该余项比 $dy$ 本身更快地消失。

To compute $\sqrt{2}$, solve $f(x) = x^2 - 2 = 0$. With $f'(x) = 2x$ the iteration simplifies to the classical averaging formula:

$$ x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right). $$

Starting from $x_0 = 1.5$:

$$ x_1 = \tfrac{1}{2}(1.5 + 1.3333\ldots) = 1.416667, \quad x_2 = 1.414216, \quad x_3 = 1.414214. $$

Three steps already match $\sqrt{2} = 1.4142136\ldots$ to six decimals, illustrating the doubling of correct digits.

要计算 $\sqrt{2}$，解 $f(x) = x^2 - 2 = 0$。由 $f'(x) = 2x$，迭代化简为经典的取平均公式：

$$ x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right). $$

从 $x_0 = 1.5$ 开始：

$$ x_1 = \tfrac{1}{2}(1.5 + 1.3333\ldots) = 1.416667, \quad x_2 = 1.414216, \quad x_3 = 1.414214. $$

三步就已与 $\sqrt{2} = 1.4142136\ldots$ 吻合到六位小数，体现了正确位数翻倍的特性。

Let $r$ be a root and expand $f$ about $x_n$ with Taylor's theorem with remainder:

$$ 0 = f(r) = f(x_n) + f'(x_n)(r - x_n) + \tfrac{1}{2}f''(\xi)(r - x_n)^2 $$

for some $\xi$ between $x_n$ and $r$. Divide by $f'(x_n)$ and rearrange, using the iteration $x_{n+1} = x_n - f(x_n)/f'(x_n)$:

$$ r - x_{n+1} = -\frac{f''(\xi)}{2 f'(x_n)}(r - x_n)^2. $$

Taking absolute values gives $|x_{n+1} - r| \le M\,|x_n - r|^2$ with $M = \max|f''| / (2 \min |f'|)$ near $r$. The squared error is the hallmark of quadratic convergence.

设 $r$ 为一个根，用带余项的泰勒定理（Taylor）把 $f$ 在 $x_n$ 处展开：

$$ 0 = f(r) = f(x_n) + f'(x_n)(r - x_n) + \tfrac{1}{2}f''(\xi)(r - x_n)^2 $$

其中 $\xi$ 介于 $x_n$ 与 $r$ 之间。两边除以 $f'(x_n)$ 并整理，代入迭代 $x_{n+1} = x_n - f(x_n)/f'(x_n)$：

$$ r - x_{n+1} = -\frac{f''(\xi)}{2 f'(x_n)}(r - x_n)^2. $$

取绝对值得到 $|x_{n+1} - r| \le M\,|x_n - r|^2$，其中在 $r$ 附近 $M = \max|f''| / (2 \min |f'|)$。误差被平方正是二次收敛的标志。

Find the root of $f(x) = \cos x - x$, which has no closed form. Here $f'(x) = -\sin x - 1$, so the iteration is

$$ x_{n+1} = x_n - \frac{\cos x_n - x_n}{-\sin x_n - 1} = x_n + \frac{\cos x_n - x_n}{\sin x_n + 1}. $$

Start from $x_0 = 1$ (a reasonable guess since $\cos 1 = 0.5403 < 1$, so the root lies below $1$):

$$ x_1 = 1 + \frac{0.5403 - 1}{0.8415 + 1} = 0.7504, \quad x_2 = 0.7391, \quad x_3 = 0.7390851, \quad x_4 = 0.7390851. $$

By the third step the iterate is stable to seven decimals at the Dottie number $0.7390851\ldots$, the unique real solution of $\cos x = x$. The fast stabilization is quadratic convergence in action.

求 $f(x) = \cos x - x$ 的根，它没有封闭形式的解。此处 $f'(x) = -\sin x - 1$，所以迭代为

$$ x_{n+1} = x_n - \frac{\cos x_n - x_n}{-\sin x_n - 1} = x_n + \frac{\cos x_n - x_n}{\sin x_n + 1}. $$

从 $x_0 = 1$ 开始（这是合理的猜测，因为 $\cos 1 = 0.5403 < 1$，所以根在 $1$ 以下）：

$$ x_1 = 1 + \frac{0.5403 - 1}{0.8415 + 1} = 0.7504, \quad x_2 = 0.7391, \quad x_3 = 0.7390851, \quad x_4 = 0.7390851. $$

到第三步，迭代值已稳定到七位小数，即 Dottie 数 $0.7390851\ldots$，它是 $\cos x = x$ 的唯一实数解。这种快速稳定正是二次收敛的体现。

Newton's method is not guaranteed to converge from every start. Two classic failures:

Cycling. For $f(x) = x^3 - 2x + 2$ starting at $x_0 = 0$: $f(0) = 2$, $f'(0) = -2$, so $x_1 = 0 - 2/(-2) = 1$. Then $f(1) = 1$, $f'(1) = 1$, so $x_2 = 1 - 1/1 = 0$. The iteration bounces $0 \to 1 \to 0 \to 1$ forever, never approaching the real root near $x = -1.77$.

Flat derivative. If some iterate lands where $f'(x_n) \approx 0$, the update $-f(x_n)/f'(x_n)$ is enormous and the next guess is flung far away. For $f(x) = x^3 - x$ near $x = 1/\sqrt3$ (an inflection-flanked critical region), a poorly chosen start can diverge. The remedy is a good initial guess, often from a sketch or a bisection step, and a check that $f'$ is not small along the way.

牛顿法不保证从任意起点都收敛。两个经典的失败例子：

循环。 对 $f(x) = x^3 - 2x + 2$，从 $x_0 = 0$ 开始：$f(0) = 2$，$f'(0) = -2$，所以 $x_1 = 0 - 2/(-2) = 1$。然后 $f(1) = 1$，$f'(1) = 1$，所以 $x_2 = 1 - 1/1 = 0$。迭代永远在 $0 \to 1 \to 0 \to 1$ 之间来回，永不接近 $x = -1.77$ 附近的实根。

导数过平。 若某次迭代落在 $f'(x_n) \approx 0$ 处，更新量 $-f(x_n)/f'(x_n)$ 会极大，下一个猜测被甩到很远。对 $f(x) = x^3 - x$，在 $x = 1/\sqrt3$ 附近（一个拐点（inflection point）旁的临界区域），选不好的起点会发散（divergence）。补救办法是给一个好的初始猜测，通常来自草图或一步二分法，并沿途检查 $f'$ 不会太小。

All three limits below have the form $\tfrac{0}{0}$ as $x \to 0$, yet they disagree, which is exactly why the form is called indeterminate:

$$ \lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}, \qquad \lim_{x \to 0} \frac{x}{x^2} = +\infty. $$

The numerator and denominator both tend to $0$ in every case, but their relative rates differ, and only the rates determine the limit.

下面三个极限当 $x \to 0$ 时都是 $\tfrac{0}{0}$ 形式，但结果各不相同，这正是它被称为不定式的原因：

$$ \lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}, \qquad \lim_{x \to 0} \frac{x}{x^2} = +\infty. $$

每种情形分子和分母都趋于 $0$，但它们的相对速率不同，而只有速率决定极限。

Not every indeterminate form needs calculus. The limit

$$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} $$

is $\tfrac{0}{0}$ on substitution, but factoring the numerator removes the trouble:

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

The form was indeterminate, but the cancellation reveals the answer is forced once the shared factor $(x-3)$ is removed. This is worth remembering: factoring, rationalizing, and using known limits such as $\lim_{x\to0}\frac{\sin x}{x}=1$ often beat L'Hopital and avoid messy derivatives.

不是每个不定式都需要微积分。极限

$$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} $$

代入后是 $\tfrac{0}{0}$，但把分子因式分解就消除了麻烦：

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

形式原本是不定的，但约分表明：一旦消去公因式 $(x-3)$，答案就被钉死了。这点值得记住：因式分解、有理化，以及利用已知极限如 $\lim_{x\to0}\frac{\sin x}{x}=1$，往往胜过洛必达法则，还能避开繁琐的求导。

Consider $\displaystyle \lim_{x \to 0^+} \frac{1}{x} \cdot e^{-1/x}$. Substituting suggests $\infty \cdot 0$, which looks indeterminate. But substitute $t = 1/x \to +\infty$:

$$ \frac{1}{x} e^{-1/x} = t\,e^{-t} = \frac{t}{e^t} \ \longrightarrow\ 0, $$

since the exponential dominates any power. The point is that the form $0 \cdot \infty$ flagged that work was needed, but once rewritten the rate comparison settles it decisively. Form names tell you where to look, not what the answer is.

考虑 $\displaystyle \lim_{x \to 0^+} \frac{1}{x} \cdot e^{-1/x}$。代入显示为 $\infty \cdot 0$，看起来像不定式。但作替换 $t = 1/x \to +\infty$：

$$ \frac{1}{x} e^{-1/x} = t\,e^{-t} = \frac{t}{e^t} \ \longrightarrow\ 0, $$

因为指数函数压倒任何幂。要点是：$0 \cdot \infty$ 这一形式提示需要做功夫，但一旦改写，速率比较就果断地定下答案。形式的名字只告诉你往哪里看，而不告诉你答案是什么。

Evaluate $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$. Substitution gives $\tfrac{0}{0}$, so differentiate top and bottom:

$$ \lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}. $$

This is again $\tfrac{0}{0}$, so apply the rule once more:

$$ = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}. $$

Each application must be justified by re-checking the form, which here remains $\tfrac{0}{0}$ until the final step.

求 $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$。代入得 $\tfrac{0}{0}$，于是对上下分别求导：

$$ \lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}. $$

这仍是 $\tfrac{0}{0}$，所以再用一次法则：

$$ = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}. $$

每次应用都必须通过重新核对形式来证明其合理性，这里形式直到最后一步之前都保持 $\tfrac{0}{0}$。

Consider the $\tfrac{0}{0}$ case at a finite $a$, and define $f(a) = g(a) = 0$ to make $f, g$ continuous there. The Cauchy Mean Value Theorem states that for $x$ near $a$ there is a point $c$ strictly between $a$ and $x$ with

$$ \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)}. $$

Since $f(a) = g(a) = 0$, the left side equals $f(x)/g(x)$, so

$$ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}. $$

As $x \to a$, the intermediate point $c$ is squeezed to $a$ as well. If $f'(x)/g'(x) \to L$, then $f'(c)/g'(c) \to L$, and therefore $f(x)/g(x) \to L$. This is the rule.

考虑在有限点 $a$ 处的 $\tfrac{0}{0}$ 情形，并定义 $f(a) = g(a) = 0$ 使 $f, g$ 在那里连续。柯西中值定理（Cauchy Mean Value Theorem）指出，对 $a$ 附近的 $x$，存在严格介于 $a$ 与 $x$ 之间的点 $c$，使

$$ \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)}. $$

由于 $f(a) = g(a) = 0$，左边等于 $f(x)/g(x)$，所以

$$ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}. $$

当 $x \to a$ 时，中间点 $c$ 也被夹向 $a$。若 $f'(x)/g'(x) \to L$，则 $f'(c)/g'(c) \to L$，从而 $f(x)/g(x) \to L$。这就是该法则。

Evaluate $\displaystyle \lim_{x \to \infty} \frac{x^2}{e^x}$, an $\tfrac{\infty}{\infty}$ form. Differentiating top and bottom:

$$ \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}, $$

still $\tfrac{\infty}{\infty}$, so apply the rule again:

$$ = \lim_{x \to \infty} \frac{2}{e^x} = 0. $$

Each pass lowers the power of $x$ by one while the exponential is untouched; after two passes the numerator is constant and the limit collapses to $0$. The same argument shows $\lim_{x\to\infty} x^n/e^x = 0$ for every fixed $n$: exponentials outgrow all polynomials.

求 $\displaystyle \lim_{x \to \infty} \frac{x^2}{e^x}$，这是 $\tfrac{\infty}{\infty}$ 形式。对上下求导：

$$ \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}, $$

仍是 $\tfrac{\infty}{\infty}$，所以再用一次法则：

$$ = \lim_{x \to \infty} \frac{2}{e^x} = 0. $$

每一次都把 $x$ 的幂降低一次，而指数函数不变；两次之后分子变成常数，极限塌缩为 $0$。同样的论证表明对每个固定的 $n$ 都有 $\lim_{x\to\infty} x^n/e^x = 0$：指数函数增长快过所有多项式。

Evaluate $\displaystyle \lim_{x \to 0^+} \frac{\sin x - x}{x^3}$. The form is $\tfrac{0}{0}$. Apply L'Hopital, re-checking the form at every step:

$$ \lim_{x\to0^+}\frac{\sin x - x}{x^3} = \lim_{x\to0^+}\frac{\cos x - 1}{3x^2} \quad (\tfrac00) = \lim_{x\to0^+}\frac{-\sin x}{6x} \quad (\tfrac00) = \lim_{x\to0^+}\frac{-\cos x}{6} = -\frac16. $$

Three applications were each justified by the $\tfrac00$ form persisting; the moment the form resolves (the last quotient is $\tfrac{-1}{6}$, not indeterminate) we stop and read off the value $-\tfrac16$. This matches the Taylor result $\sin x - x = -x^3/6 + \cdots$ from Section 7.

求 $\displaystyle \lim_{x \to 0^+} \frac{\sin x - x}{x^3}$。形式是 $\tfrac{0}{0}$。应用洛必达法则，每步都重新核对形式：

$$ \lim_{x\to0^+}\frac{\sin x - x}{x^3} = \lim_{x\to0^+}\frac{\cos x - 1}{3x^2} \quad (\tfrac00) = \lim_{x\to0^+}\frac{-\sin x}{6x} \quad (\tfrac00) = \lim_{x\to0^+}\frac{-\cos x}{6} = -\frac16. $$

三次应用都因 $\tfrac00$ 形式持续存在而成立；一旦形式化解（最后的商是 $\tfrac{-1}{6}$，不再是不定式），我们就停下并读出值 $-\tfrac16$。这与第 7 节的泰勒（Taylor）结果 $\sin x - x = -x^3/6 + \cdots$ 一致。

Evaluate $\displaystyle \lim_{x \to 0^+} x \ln x$, which has the form $0 \cdot (-\infty)$. Rewrite as a quotient so the rule applies:

$$ \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} \quad (\text{form } \tfrac{-\infty}{\infty}). $$ $$ = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0. $$

Choosing to put $\ln x$ on top was deliberate: it differentiates to a simple $1/x$, which cancels cleanly.

求 $\displaystyle \lim_{x \to 0^+} x \ln x$，它是 $0 \cdot (-\infty)$ 形式。改写成商，使法则可用：

$$ \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} \quad (\text{形式 } \tfrac{-\infty}{\infty}). $$ $$ = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0. $$

把 $\ln x$ 放在分子是有意为之：它求导后得到简单的 $1/x$，能干净地约去。

Evaluate $\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x}$, a $1^\infty$ form. Let $y$ denote the expression and take logarithms:

$$ \ln y = x \ln\!\left(1 + \frac{1}{x}\right) = \frac{\ln(1 + 1/x)}{1/x} \quad (\text{form } \tfrac{0}{0}). $$

Let $t = 1/x \to 0^+$ and apply L'Hopital to $\dfrac{\ln(1 + t)}{t}$:

$$ \lim_{t \to 0^+} \frac{\ln(1 + t)}{t} = \lim_{t \to 0^+} \frac{1/(1+t)}{1} = 1. $$

So $\ln y \to 1$, and therefore $y \to e^1 = e$. This recovers the classical definition of $e$.

求 $\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x}$，这是 $1^\infty$ 形式。设 $y$ 表示这个表达式并取对数：

$$ \ln y = x \ln\!\left(1 + \frac{1}{x}\right) = \frac{\ln(1 + 1/x)}{1/x} \quad (\text{形式 } \tfrac{0}{0}). $$

令 $t = 1/x \to 0^+$，对 $\dfrac{\ln(1 + t)}{t}$ 应用洛必达法则：

$$ \lim_{t \to 0^+} \frac{\ln(1 + t)}{t} = \lim_{t \to 0^+} \frac{1/(1+t)}{1} = 1. $$

所以 $\ln y \to 1$，从而 $y \to e^1 = e$。这就还原了 $e$ 的经典定义。

Evaluate $\displaystyle \lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{\sin x} \right)$, an $\infty - \infty$ form. Combine over a common denominator to make a single quotient:

$$ \frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x} \quad (\text{form } \tfrac00). $$

Now apply L'Hopital twice, or use $\sin x = x - x^3/6 + \cdots$ and $x\sin x = x^2 + \cdots$:

$$ \frac{\sin x - x}{x \sin x} = \frac{-x^3/6 + \cdots}{x^2 + \cdots} = \frac{-x/6 + \cdots}{1 + \cdots} \ \longrightarrow\ 0. $$

The difference of two blow-ups is finite, in fact zero, because the singular parts $1/x$ cancel and only a vanishing remainder survives.

求 $\displaystyle \lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{\sin x} \right)$，这是 $\infty - \infty$ 形式。通分化为单个商：

$$ \frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x} \quad (\text{形式 } \tfrac00). $$

现在用两次洛必达法则，或利用 $\sin x = x - x^3/6 + \cdots$ 和 $x\sin x = x^2 + \cdots$：

$$ \frac{\sin x - x}{x \sin x} = \frac{-x^3/6 + \cdots}{x^2 + \cdots} = \frac{-x/6 + \cdots}{1 + \cdots} \ \longrightarrow\ 0. $$

两个发散量之差是有限的，事实上为零，因为奇异部分 $1/x$ 相互抵消，只剩下一个趋于零的余项。

Evaluate $\displaystyle \lim_{x \to \infty} x^{1/x}$, an $\infty^0$ form. Set $y = x^{1/x}$ and take logarithms:

$$ \ln y = \frac{\ln x}{x} \quad (\text{form } \tfrac{\infty}{\infty}). $$

By L'Hopital, $\displaystyle \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0$, so $\ln y \to 0$ and therefore $y \to e^0 = 1$. A base growing to infinity raised to a power shrinking to zero can balance exactly to $1$; the logarithm makes the competition explicit.

求 $\displaystyle \lim_{x \to \infty} x^{1/x}$，这是 $\infty^0$ 形式。令 $y = x^{1/x}$ 并取对数：

$$ \ln y = \frac{\ln x}{x} \quad (\text{形式 } \tfrac{\infty}{\infty}). $$

由洛必达法则，$\displaystyle \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0$，所以 $\ln y \to 0$，从而 $y \to e^0 = 1$。底数增长到无穷、指数缩小到零，二者可以恰好平衡为 $1$；对数把这场较量显式地呈现出来。

Evaluate $\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x^3}$. Repeated L'Hopital works, but the Taylor series is cleaner. Using $\sin x = x - \tfrac{x^3}{6} + \tfrac{x^5}{120} - \cdots$,

$$ x - \sin x = \frac{x^3}{6} - \frac{x^5}{120} + \cdots, $$ $$ \frac{x - \sin x}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots \ \longrightarrow\ \frac{1}{6}. $$

The leading nonzero term of the numerator is cubic, which is why the limit over $x^3$ is finite and equal to $\tfrac{1}{6}$.

求 $\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x^3}$。反复用洛必达法则也行，但泰勒级数（Taylor series）更简洁。利用 $\sin x = x - \tfrac{x^3}{6} + \tfrac{x^5}{120} - \cdots$，

$$ x - \sin x = \frac{x^3}{6} - \frac{x^5}{120} + \cdots, $$ $$ \frac{x - \sin x}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots \ \longrightarrow\ \frac{1}{6}. $$

分子第一个非零项是三次的，这就是为什么除以 $x^3$ 后极限有限且等于 $\tfrac{1}{6}$。

The rule requires that $\lim f'/g'$ exist. If it does not, the rule is silent; the original limit may still exist. Consider

$$ \lim_{x \to \infty} \frac{x + \sin x}{x}. $$

This is $\tfrac{\infty}{\infty}$, but differentiating gives $\dfrac{1 + \cos x}{1}$, which oscillates and has no limit. L'Hopital yields nothing. Yet dividing directly,

$$ \frac{x + \sin x}{x} = 1 + \frac{\sin x}{x} \ \longrightarrow\ 1, $$

since $\sin x / x \to 0$. The lesson: a failed L'Hopital attempt is not a proof that the limit fails to exist. Always have algebraic methods in reserve.

该法则要求 $\lim f'/g'$ 存在。若它不存在，法则就保持沉默；原极限仍可能存在。考虑

$$ \lim_{x \to \infty} \frac{x + \sin x}{x}. $$

这是 $\tfrac{\infty}{\infty}$，但求导得到 $\dfrac{1 + \cos x}{1}$，它在振荡、没有极限。洛必达法则给不出结果。然而直接相除，

$$ \frac{x + \sin x}{x} = 1 + \frac{\sin x}{x} \ \longrightarrow\ 1, $$

因为 $\sin x / x \to 0$。教训是：洛必达法则尝试失败并不能证明极限不存在。手边永远要备有代数方法。

Return to $\sqrt{4.1}$ from Section 1, now with the second-order term. With $f(x) = \sqrt x$ at $a = 4$, $f'(x) = \tfrac12 x^{-1/2}$ and $f''(x) = -\tfrac14 x^{-3/2}$, so $f''(4) = -\tfrac14 \cdot \tfrac18 = -\tfrac{1}{32}$. The quadratic model is

$$ f(x) \approx 2 + \tfrac14(x-4) + \tfrac{1}{2}\!\left(-\tfrac{1}{32}\right)(x-4)^2. $$

At $x = 4.1$, $(x-4)^2 = 0.01$:

$$ \sqrt{4.1} \approx 2 + 0.025 - \tfrac{1}{64}(0.01) = 2.025 - 0.00015625 = 2.0248438. $$

Compared with the true $2.0248457\ldots$, the quadratic estimate is correct to seven decimals, versus four for the linear one. The negative $f''$ correctly pulls the overestimate back down, confirming the sign analysis: concave-down curves sit below their tangent lines.

回到第 1 节的 $\sqrt{4.1}$，这次带上二阶项。取 $f(x) = \sqrt x$，在 $a = 4$ 处，$f'(x) = \tfrac12 x^{-1/2}$，$f''(x) = -\tfrac14 x^{-3/2}$，所以 $f''(4) = -\tfrac14 \cdot \tfrac18 = -\tfrac{1}{32}$。二次模型为

$$ f(x) \approx 2 + \tfrac14(x-4) + \tfrac{1}{2}\!\left(-\tfrac{1}{32}\right)(x-4)^2. $$

在 $x = 4.1$ 处，$(x-4)^2 = 0.01$：

$$ \sqrt{4.1} \approx 2 + 0.025 - \tfrac{1}{64}(0.01) = 2.025 - 0.00015625 = 2.0248438. $$

与真值 $2.0248457\ldots$ 相比，二次估计精确到七位小数，而线性估计只到四位。负的 $f''$ 正确地把高估值往下拉，印证了符号分析：向下凹的曲线位于切线下方。

Evaluate $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x - \tfrac{x^2}{2}}{x^3}$. Repeated L'Hopital would require three differentiations of a growing expression; the series is immediate. Using $e^x = 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6} + \cdots$,

$$ e^x - 1 - x - \tfrac{x^2}{2} = \frac{x^3}{6} + \frac{x^4}{24} + \cdots, $$ $$ \frac{e^x - 1 - x - \tfrac{x^2}{2}}{x^3} = \frac16 + \frac{x}{24} + \cdots \ \longrightarrow\ \frac16. $$

The lesson generalizes: when a limit is built from standard functions near $0$, expanding numerator and denominator to the first surviving power is usually faster and less error-prone than iterating L'Hopital, and it exposes why the answer is what it is.

求 $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x - \tfrac{x^2}{2}}{x^3}$。反复用洛必达法则需要对一个越来越长的表达式求三次导；用级数则立竿见影。利用 $e^x = 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6} + \cdots$，

$$ e^x - 1 - x - \tfrac{x^2}{2} = \frac{x^3}{6} + \frac{x^4}{24} + \cdots, $$ $$ \frac{e^x - 1 - x - \tfrac{x^2}{2}}{x^3} = \frac16 + \frac{x}{24} + \cdots \ \longrightarrow\ \frac16. $$

这个经验可以推广：当一个极限由 $0$ 附近的标准函数构成时，把分子和分母展开到第一个不消失的幂，通常比反复用洛必达法则更快、更不易出错，并且能揭示答案为什么是这样。