Linear Approximation and L'Hopital's Rule · Solutions线性近似与洛必达法则 · 解答

(a) Apply L'Hopital twice to a 0/0 form(a) 对 $0/0$ 型两次使用洛必达法则 M1·M1·A1

Direct substitution gives $\tfrac{0}{0}$. (M1) Apply L'Hopital once: numerator becomes $e^{x}-1$, denominator $2x$. At $x=0$ this is again $\tfrac{0}{0}$. (M1) Apply L'Hopital a second time:直接代入得 $\tfrac{0}{0}$。(M1) 使用一次洛必达法则：分子变为 $e^{x}-1$，分母为 $2x$。在 $x=0$ 处仍为 $\tfrac{0}{0}$。(M1) 再次使用洛必达法则：

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

(A1)

(b) Apply L'Hopital three times to an infty/infty form(b) 对 $\infty/\infty$ 型三次使用洛必达法则 M1·M1·A1

The form is $\infty/\infty$. Differentiate numerator and denominator: (M1)极限形式为 $\infty/\infty$。对分子分母分别求导：(M1)

$$ \frac{x^{3}}{e^{x}}\xrightarrow{L'H}\frac{3x^{2}}{e^{x}}\xrightarrow{L'H}\frac{6x}{e^{x}}\xrightarrow{L'H}\frac{6}{e^{x}}\longrightarrow 0. $$

Each step maintains the $\infty/\infty$ form until the last, where the limit is $6/\infty=0$. (M1·A1)每步均保持 $\infty/\infty$ 形式，直至最后一步极限为 $6/\infty=0$。(M1·A1)

(c) Apply L'Hopital once, then use the standard limit(c) 使用一次洛必达法则，再利用标准极限 M1·A1

The form is $\tfrac{0}{0}$ at $x=0$. Differentiate: $\dfrac{\sec^{2}x-1}{3x^{2}}=\dfrac{\tan^{2}x}{3x^{2}}$. This is still $\tfrac{0}{0}$. (M1) Write在 $x=0$ 处形式为 $\tfrac{0}{0}$。求导得 $\dfrac{\sec^{2}x-1}{3x^{2}}=\dfrac{\tan^{2}x}{3x^{2}}$，仍为 $\tfrac{0}{0}$。(M1) 写成

$$ \frac{\tan^{2}x}{3x^{2}}=\frac{1}{3}\left(\frac{\tan x}{x}\right)^{2}\longrightarrow\frac{1}{3}\cdot 1^{2}=\frac{1}{3}, $$

since $\lim_{x\to 0}\dfrac{\tan x}{x}=1$. (A1)因为 $\lim_{x\to 0}\dfrac{\tan x}{x}=1$。(A1)

(a) 0 times infty: rewrite as a quotient(a) $0\cdot\infty$ 型：改写为商 M1·A1

The form is $0\cdot(-\infty)$. Rewrite as $\dfrac{\ln x}{1/x}$, which is $\dfrac{-\infty}{\infty}$. (M1) Apply L'Hopital:极限形式为 $0\cdot(-\infty)$。改写为 $\dfrac{\ln x}{1/x}$，即 $\dfrac{-\infty}{\infty}$ 型。(M1) 使用洛必达法则：

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

(A1)

(b) NOT indeterminate: L'Hopital does not apply(b) 非不定式：洛必达法则不适用 M1·A1

As $x\to 0$: the numerator $\sin x\to 0$ but the denominator $x+1\to 1\ne 0$. (M1) The form is $0/1$, which is not indeterminate. L'Hopital's rule requires the form to be $0/0$ or $\pm\infty/\pm\infty$; neither holds here.当 $x\to 0$ 时：分子 $\sin x\to 0$，但分母 $x+1\to 1\ne 0$。(M1) 形式为 $0/1$，不是不定式。洛必达法则要求形式为 $0/0$ 或 $\pm\infty/\pm\infty$，两者在此均不成立。

$$ \lim_{x\to 0}\frac{\sin x}{x+1}=\frac{\sin 0}{0+1}=\frac{0}{1}=0. $$

(A1)

(c) infty minus infty: rationalise to a quotient(c) $\infty-\infty$ 型：有理化为商 M1·A1

The form is $\infty-\infty$. Multiply by the conjugate: (M1)极限形式为 $\infty-\infty$。乘以共轭表达式：(M1)

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

As $x\to\infty$, $1/x\to 0$, so the expression tends to $\dfrac{1}{\sqrt{1}+1}=\dfrac{1}{2}$. (A1)当 $x\to\infty$ 时，$1/x\to 0$，表达式趋向 $\dfrac{1}{\sqrt{1}+1}=\dfrac{1}{2}$。(A1)

(a) Full statement of L'Hopital's rule(a) 洛必达法则的完整表述 R1·R1·A1

Hypotheses.假设条件。 Suppose $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ itself), that $g'(x)\ne 0$ for $x$ near $a$ but $x\ne a$, and that $\displaystyle\lim_{x\to a}f(x)=0$ and $\displaystyle\lim_{x\to a}g(x)=0$. (R1)设 $f$ 和 $g$ 在含 $a$ 的开区间上可微（$a$ 本身可除外），$g'(x)\ne 0$ 在 $a$ 附近（但 $x\ne a$）成立，且 $\displaystyle\lim_{x\to a}f(x)=0$，$\displaystyle\lim_{x\to a}g(x)=0$。(R1)

Suppose further that the limit $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists (or equals $\pm\infty$). (R1)进一步设极限 $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$ 存在（或等于 $\pm\infty$）。(R1)

Conclusion.结论。 Then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$. (A1)则 $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$。(A1)

(b) Why the application is invalid, and correct evaluation(b) 为何该应用无效，以及正确求值 R1·M1·A1·M1·A1

The form as $x\to 0$ is indeed $0/0$. However, to apply L'Hopital, we need $\displaystyle\lim_{x\to 0}\frac{[x^{2}\sin(1/x)]'}{[\sin x]'}$ to exist. (R1)当 $x\to 0$ 时，极限形式确为 $0/0$。然而，使用洛必达法则需要 $\displaystyle\lim_{x\to 0}\frac{[x^{2}\sin(1/x)]'}{[\sin x]'}$ 存在。(R1)

Compute the derivative of the numerator: $[x^{2}\sin(1/x)]'=2x\sin(1/x)+x^{2}\cos(1/x)\cdot(-1/x^{2})=2x\sin(1/x)-\cos(1/x)$. (M1)计算分子的导数：$[x^{2}\sin(1/x)]'=2x\sin(1/x)+x^{2}\cos(1/x)\cdot(-1/x^{2})=2x\sin(1/x)-\cos(1/x)$。(M1)

As $x\to 0$, the term $\cos(1/x)$ oscillates between $-1$ and $1$ without settling to any limit. Therefore $\displaystyle\lim_{x\to 0}[x^{2}\sin(1/x)]'$ does not exist, and the hypothesis of L'Hopital's rule fails. The rule cannot be applied. (A1)当 $x\to 0$ 时，$\cos(1/x)$ 在 $-1$ 和 $1$ 之间振荡，不趋于任何极限。因此 $\displaystyle\lim_{x\to 0}[x^{2}\sin(1/x)]'$ 不存在，洛必达法则的假设条件不满足，该法则不可使用。(A1)

Correct method.正确方法。 Write the limit as a product and use the bound $|\sin(1/x)|\le 1$: (M1)将极限写为乘积，利用界 $|\sin(1/x)|\le 1$：(M1)

$$ \frac{x^{2}\sin(1/x)}{\sin x}=\underbrace{\frac{x}{\sin x}}_{\to\, 1}\cdot\underbrace{x\sin(1/x)}_{\to\, 0}, $$

since $x\sin(1/x)\to 0$ by the squeeze theorem ($|x\sin(1/x)|\le|x|\to 0$) and $x/\sin x\to 1$. The limit is $1\cdot 0=0$. (A1)因为由夹逼定理（$|x\sin(1/x)|\le|x|\to 0$）得 $x\sin(1/x)\to 0$，且 $x/\sin x\to 1$。极限为 $1\cdot 0=0$。(A1)

(a) x^x as x approaches 0+: form 0^0(a) $x\to 0^+$ 时 $x^x$：$0^0$ 型 M1·A1·A1

Let $y=x^{x}$, so $\ln y=x\ln x$. This is the $0\cdot(-\infty)$ form. Rewrite as $\dfrac{\ln x}{1/x}$. (M1)令 $y=x^{x}$，则 $\ln y=x\ln x$，为 $0\cdot(-\infty)$ 型。改写为 $\dfrac{\ln x}{1/x}$。(M1)

$$ \lim_{x\to 0^{+}}\frac{\ln x}{1/x}\overset{L'H}{=}\lim_{x\to 0^{+}}\frac{1/x}{-1/x^{2}}=\lim_{x\to 0^{+}}(-x)=0. $$

Therefore $\lim_{x\to 0^{+}}\ln y=0$, so $\lim_{x\to 0^{+}}y=e^{0}=1$. (A1·A1)故 $\lim_{x\to 0^{+}}\ln y=0$，从而 $\lim_{x\to 0^{+}}y=e^{0}=1$。(A1·A1)

(b) x^{1/x} as x approaches infty: form infty^0(b) $x\to\infty$ 时 $x^{1/x}$：$\infty^0$ 型 M1·A1·A1

Let $y=x^{1/x}$, so $\ln y=\dfrac{\ln x}{x}$. This is $\infty/\infty$. (M1)令 $y=x^{1/x}$，则 $\ln y=\dfrac{\ln x}{x}$，为 $\infty/\infty$ 型。(M1)

$$ \lim_{x\to\infty}\frac{\ln x}{x}\overset{L'H}{=}\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac{1}{x}=0. $$

Therefore $\lim_{x\to\infty}\ln y=0$, so $\lim_{x\to\infty}y=e^{0}=1$. (A1·A1)故 $\lim_{x\to\infty}\ln y=0$，从而 $\lim_{x\to\infty}y=e^{0}=1$。(A1·A1)

(c) (1 + 3/x)^x as x approaches infty: form 1^infty(c) $x\to\infty$ 时 $(1+3/x)^x$：$1^\infty$ 型 M1·M1·A1·A1

Let $y=\left(1+\dfrac{3}{x}\right)^{x}$, so $\ln y=x\ln\!\left(1+\dfrac{3}{x}\right)$. This is $\infty\cdot 0$. Rewrite: (M1)令 $y=\left(1+\dfrac{3}{x}\right)^{x}$，则 $\ln y=x\ln\!\left(1+\dfrac{3}{x}\right)$，为 $\infty\cdot 0$ 型。改写：(M1)

$$ \ln y=\frac{\ln(1+3/x)}{1/x}. $$

The form is $0/0$ as $x\to\infty$. Apply L'Hopital: (M1)当 $x\to\infty$ 时形式为 $0/0$。使用洛必达法则：(M1)

$$ \lim_{x\to\infty}\frac{\ln(1+3/x)}{1/x}\overset{L'H}{=}\lim_{x\to\infty}\frac{\dfrac{-3/x^{2}}{1+3/x}}{-1/x^{2}}=\lim_{x\to\infty}\frac{3}{1+3/x}=3. $$

Therefore $\lim_{x\to\infty}\ln y=3$, so $\lim_{x\to\infty}y=e^{3}$. (A1·A1)故 $\lim_{x\to\infty}\ln y=3$，从而 $\lim_{x\to\infty}y=e^{3}$。(A1·A1)

(a) Two Newton iterations(a) 两次牛顿迭代 M1·A1·M1·A1

$f(x)=x^{2}-3$, $f'(x)=2x$. The iteration is $x_{n+1}=x_{n}-\dfrac{x_{n}^{2}-3}{2x_{n}}$. (M1)$f(x)=x^{2}-3$，$f'(x)=2x$。迭代公式为 $x_{n+1}=x_{n}-\dfrac{x_{n}^{2}-3}{2x_{n}}$。(M1)

$x_{1}=1-\dfrac{1-3}{2}=1+1=2.0000$. (A1)$x_{1}=1-\dfrac{1-3}{2}=1+1=2.0000$。(A1)

(M1) $x_{2}=2-\dfrac{4-3}{4}=2-0.25=1.7500$. (A1) (True value: $\sqrt{3}\approx 1.7321$.)(M1) $x_{2}=2-\dfrac{4-3}{4}=2-0.25=1.7500$。(A1)（真实值：$\sqrt{3}\approx 1.7321$。）

(b) Geometric interpretation(b) 几何解释 R1·A1

At each step, draw the tangent line to $y=f(x)$ at the current point $(x_{n},f(x_{n}))$. (R1) The next iterate $x_{n+1}$ is the $x$-intercept of that tangent line: solving $0=f(x_{n})+f'(x_{n})(x-x_{n})$ gives $x=x_{n}-f(x_{n})/f'(x_{n})$. (A1)每步在当前点 $(x_{n},f(x_{n}))$ 处作 $y=f(x)$ 的切线。(R1) 下一个迭代值 $x_{n+1}$ 是该切线与 $x$ 轴的交点：解方程 $0=f(x_{n})+f'(x_{n})(x-x_{n})$ 得 $x=x_{n}-f(x_{n})/f'(x_{n})$。(A1)

(c) Condition for failure(c) 失效条件 R1·A1

If $f'(x_{n})\approx 0$ (the function is nearly flat at the current iterate), the tangent line is nearly horizontal and its $x$-intercept is enormously far from $x_{n}$. (R1) For example, applying Newton's method to $f(x)=x^{3}-x$ starting near the inflection point $x=0$ can send the iterate far from any root. (A1)若 $f'(x_{n})\approx 0$（函数在当前迭代点附近近乎水平），则切线几乎水平，其与 $x$ 轴的交点距 $x_{n}$ 极远。(R1) 例如，对 $f(x)=x^{3}-x$ 从接近拐点 $x=0$ 处出发使用牛顿迭代法，迭代值可能远离任何根。(A1)

(a) Three applications of L'Hopital(a) 三次使用洛必达法则 M1·M1·M1·A1

At $x=0$, numerator and denominator are both $0$ (form $0/0$). (M1) First application:在 $x=0$ 处，分子和分母均为 $0$（$0/0$ 型）。(M1) 第一次使用：

$$ \frac{e^{x}-1-x}{3x^{2}}\quad\text{still }0/0\text{ at }x=0. $$

(M1) Second application:(M1) 第二次使用：

$$ \frac{e^{x}-1}{6x}\quad\text{still }0/0\text{ at }x=0. $$

(M1) Third application:(M1) 第三次使用：

$$ \frac{e^{x}}{6}\longrightarrow\frac{e^{0}}{6}=\frac{1}{6}. $$

(A1)

(b) Two L'Hopital applications: re-verify at each step(b) 两次洛必达法则：每步重新验证 M1·M1·A1

Form is $0/0$ at $x=0$. (M1) First step: $\dfrac{\sin x}{2x}$. This is still $0/0$ at $x=0$. (M1) Second step:在 $x=0$ 处为 $0/0$ 型。(M1) 第一步：$\dfrac{\sin x}{2x}$，在 $x=0$ 处仍为 $0/0$。(M1) 第二步：

$$ \frac{\cos x}{2}\longrightarrow\frac{1}{2}. $$

(A1)

(c) L'Hopital loops: use direct algebra instead(c) 洛必达法则陷入循环：改用直接代数化简 R1·M1·A1

The form as $x\to\infty$ is $\infty/\infty$. Differentiating: numerator becomes $e^{x}-e^{-x}$, denominator $e^{x}+e^{-x}$. This is the original quotient with numerator and denominator swapped, so applying L'Hopital again returns to $(e^{x}+e^{-x})/(e^{x}-e^{-x})$. The rule loops indefinitely without resolving the limit. (R1)当 $x\to\infty$ 时形式为 $\infty/\infty$。求导后：分子变为 $e^{x}-e^{-x}$，分母为 $e^{x}+e^{-x}$。这是原商的分子分母互换，再次使用洛必达法则将回到 $(e^{x}+e^{-x})/(e^{x}-e^{-x})$。法则无限循环，无法求出极限。(R1)

Instead, divide numerator and denominator by $e^{x}$: (M1)改为将分子分母均除以 $e^{x}$：(M1)

$$ \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}\longrightarrow\frac{1+0}{1-0}=1. $$

(A1)

(a) Dominance of e^x over every polynomial(a) $e^x$ 对所有多项式的支配性 M1·M1·A1·A1

The form is $\infty/\infty$. Apply L'Hopital once to reduce the power of $x$: (M1)极限形式为 $\infty/\infty$。使用一次洛必达法则以降低 $x$ 的幂次：(M1)

$$ \frac{x^{n}}{e^{x}}\overset{L'H}{\longrightarrow}\frac{nx^{n-1}}{e^{x}}\overset{L'H}{\longrightarrow}\frac{n(n-1)x^{n-2}}{e^{x}}\overset{L'H}{\longrightarrow}\cdots $$

Each application reduces the exponent of $x$ by $1$. The form remains $\infty/\infty$ as long as the exponent in the numerator is positive. (M1) After exactly $n$ applications, the numerator becomes the constant $n!$:每次使用将 $x$ 的指数降低 $1$。只要分子中的指数为正，形式始终保持 $\infty/\infty$。(M1) 恰好 $n$ 次使用后，分子变为常数 $n!$：

$$ \lim_{x\to\infty}\frac{n!}{e^{x}}=0, $$

since $e^{x}\to\infty$. (A1) Therefore $\lim_{x\to\infty}x^{n}/e^{x}=0$ for every positive integer $n$. (A1)因为 $e^{x}\to\infty$。(A1) 故对每个正整数 $n$，$\lim_{x\to\infty}x^{n}/e^{x}=0$。(A1)

(b) infty minus infty: combine over a common denominator(b) $\infty-\infty$ 型：通分 M1·M1·A1

Combine the two fractions: (M1)将两个分数合并：(M1)

$$ \frac{1}{\sin x}-\frac{1}{x}=\frac{x-\sin x}{x\sin x}. $$

As $x\to 0^{+}$, both numerator and denominator tend to $0$ (form $0/0$). Apply L'Hopital: (M1)当 $x\to 0^{+}$ 时，分子和分母均趋向 $0$（$0/0$ 型）。使用洛必达法则：(M1)

$$ \frac{1-\cos x}{\sin x+x\cos x}\overset{L'H}{=}\frac{\sin x}{2\cos x-x\sin x}\longrightarrow\frac{0}{2-0}=0. $$

(A1)

(c) Logarithm law avoids L'Hopital(c) 利用对数法则，无需洛必达法则 M1·A1·R1

Use the logarithm subtraction rule: (M1)利用对数相减法则：(M1)

$$ \ln(x+1)-\ln x=\ln\!\left(\frac{x+1}{x}\right)=\ln\!\left(1+\frac{1}{x}\right). $$

As $x\to\infty$, $1/x\to 0$, so $\ln(1+1/x)\to\ln 1=0$. (A1)当 $x\to\infty$ 时，$1/x\to 0$，故 $\ln(1+1/x)\to\ln 1=0$。(A1)

This is consistent with the growth-rate hierarchy: $\ln x$ grows so slowly that the difference between consecutive values $\ln(x+1)$ and $\ln x$ shrinks to zero, even though each value diverges. (R1)这与增长率层级一致：$\ln x$ 增长极为缓慢，以至于相邻值 $\ln(x+1)$ 与 $\ln x$ 之差趋向零，尽管每个值本身趋向无穷。(R1)

(a) Limit of (1+x)^{1/x} as x approaches 0+: logarithm technique(a) $x\to 0^+$ 时 $(1+x)^{1/x}$ 的极限：对数技巧 M1·M1·A1·A1

Let $y=(1+x)^{1/x}$, so $\ln y=\dfrac{\ln(1+x)}{x}$. (M1) As $x\to 0^{+}$, both numerator and denominator tend to $0$ (form $0/0$). Apply L'Hopital: (M1)令 $y=(1+x)^{1/x}$，则 $\ln y=\dfrac{\ln(1+x)}{x}$。(M1) 当 $x\to 0^{+}$ 时，分子分母均趋向 $0$（$0/0$ 型）。使用洛必达法则：(M1)

$$ \lim_{x\to 0^{+}}\frac{\ln(1+x)}{x}\overset{L'H}{=}\lim_{x\to 0^{+}}\frac{1/(1+x)}{1}=\frac{1}{1+0}=1. $$

Therefore $\lim_{x\to 0^{+}}\ln y=1$, so $\lim_{x\to 0^{+}}y=e^{1}=e$. (A1·A1)故 $\lim_{x\to 0^{+}}\ln y=1$，从而 $\lim_{x\to 0^{+}}y=e^{1}=e$。(A1·A1)

(b) Intuitive explanation via linearization of ln(b) 通过对数的线性化给出直观解释 M1·A1·R1

Linearize $g(t)=\ln(1+t)$ at $t=0$: $g'(t)=\dfrac{1}{1+t}$, so $g'(0)=1$. (M1) The linearization is $L(t)=0+1\cdot t=t$, giving the approximation $\ln(1+t)\approx t$ for small $t$. (A1)将 $g(t)=\ln(1+t)$ 在 $t=0$ 处线性化：$g'(t)=\dfrac{1}{1+t}$，故 $g'(0)=1$。(M1) 线性化为 $L(t)=0+1\cdot t=t$，给出小 $t$ 时的近似 $\ln(1+t)\approx t$。(A1)

Applying this with $t=x$ small:将此应用于小 $t=x$：

$$ \ln y=\frac{\ln(1+x)}{x}\approx\frac{x}{x}=1. $$

Therefore $y\approx e^{1}=e$. This gives an intuitive understanding of why $e$ is defined by this limit: as $x\to 0^{+}$, the linearization of $\ln$ makes the exponent of $e$ approach exactly $1$. (R1)故 $y\approx e^{1}=e$。这直观地解释了为何 $e$ 由此极限定义：当 $x\to 0^{+}$ 时，对数的线性化使 $e$ 的指数恰好趋向 $1$。(R1)

(c) Comparing linear and quadratic error at x = 0.2(c) 在 $x=0.2$ 处比较线性与二次误差 M1·A1·A1

Use $e^{0.2}\approx 1.2214$ (given). (M1)使用 $e^{0.2}\approx 1.2214$（已知）。(M1)

Linear approximation: $L(0.2)=1+0.2=1.2000$. Absolute error: $|1.2214-1.2000|=0.0214$. (A1)线性近似：$L(0.2)=1+0.2=1.2000$。绝对误差：$|1.2214-1.2000|=0.0214$。(A1)

Quadratic approximation: $Q(0.2)=1+0.2+\tfrac{1}{2}(0.04)=1+0.2+0.02=1.2200$. Absolute error: $|1.2214-1.2200|=0.0014$. (A1)二次近似：$Q(0.2)=1+0.2+\tfrac{1}{2}(0.04)=1+0.2+0.02=1.2200$。绝对误差：$|1.2214-1.2200|=0.0014$。(A1)

The ratio of errors is $0.0214/0.0014\approx 15.3$, so the quadratic approximation is roughly $15$ times more accurate at $x=0.2$.误差之比为 $0.0214/0.0014\approx 15.3$，故在 $x=0.2$ 处，二次近似的精度约为线性近似的 $15$ 倍。