陶哲轩Analysis I习题的参考解答及思考（第10章）

第10章

版本

Analysis I（第3版）。

附注

注意下，作者在本章中，集合间差集的符号由\( X \setminus Y \)改为\( X - Y \)了。

章节10.1

练习10.1.1

题目：

Suppose that \( X \) is a subset of \( \mathbf{R} \), \( x_0 \) is a limit point of \( X \), and \( f: X \to \mathbf{R} \) is a function which is differentiable at \( x_0 \) . Let \( Y \subseteq X \) be such that \( x_0 \in Y \), and \( x_0 \) is also a limit point of \( Y \). Prove that the restricted function \( f|_Y: Y \to \mathbf{R} \) is also differentiable at \( x_0 \), and has the same derivative as f at \( x_0 \). Explain why this does not contradict the discussion in Remark 10.1.2.

Remark 10.1.2的内容：

Note that we need \( x_0 \) to be a limit point in order for \( x_0 \) to be adherent to \( X - \{ x_0 \} \), otherwise the limit \( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \) would automatically be undefined. In particular, we do not define the derivative of a function at an isolated point; for instance, if one restricts the function \( f: \mathbf{R} \to \mathbf{R} \) defined by \( f(x) := x^2 \) to the domain \( X := [1, 2] \cup \{ 3 \} \), then the restriction of the function ceases to be differentiable at \( 3 \). (See however Exercise 10.1.1 below.) In practice, the domain \( X \) will almost always be an interval, and so by Lemma 9.1.21 all elements \( x_0 \) of \( X \) will automatically be limit points and we will not have to care much about these issues.

证明：

因为\( f \)在\( x_0 \)可微，因此 \( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \) 存在且等于某个数\( L \)，可得\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta, |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)，而由\( Y \subseteq X \)，可得\( Y - \{ x_0 \} \subseteq X - \{ x_0 \} \)，因此\( \forall x \in Y - \{ x_0 \}, |x - x_0| \leq \delta \)，有\( x \in X - \{ x_0 \}, |x - x_0| \leq \delta \)，进而有\( |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)。简而言之，\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in Y - \{ x_0 \}, |x - x_0| \leq \delta, |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)，即\( \lim_{x \to x_0; x \in Y - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = L = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \)。

证毕。

解释该命题为什么没有和Remark 10.1.2中的讨论矛盾：

Remark 10.1.2的函数\( f \)在限制定义域到\( [1, 2] \cup \{ 3 \} \)后， \( 3 \)不再是定义域的极限点，因此根据定义10.1.1直接不可微，然而本命题中，要求\( x_0 \)仍是子集\( Y \)的极限点，进而仍是\( Y - \{ x_0 \} \)的附着点，此时根据定义9.3.6， \( \lim_{x \to x_0; x \in Y - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \)才可能有意义（极限存在的话），因此不矛盾。

练习10.1.2

题目：

Prove Proposition 10.1.7. (Hint: the cases \( x = x_0 \) and \( x \neq x_0 \) have to be treated separately.)

Proposition 10.1.7的内容：

(Newton’s approximation). Let \( X \) be a subset of \( \mathbf{R} \), let \( x_0 \in X \) be a limit point of \( X \), let \( f: X \to \mathbf{R} \) be a function, and let \( L \) be a real number. Then the following statements are logically equivalent:

1. \( f \) is differentiable at \( x_0 \) on \( X \) with derivative \( L \).
2. For every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that \( f(x) \) is \( \epsilon|x - x_0| \)-close to \( f(x_0) + L(x - x_0 ) \) whenever \( x \in X \) is \( \delta \)-close to \( x_0 \), i.e., we have \( |f(x) - (f(x_0) + L(x - x_0))| \leq \epsilon|x - x_0| \) whenever \( x \in X \) and \( |x - x_0| \leq \delta \).

注：

当\( \epsilon \)取的越小时，\( \epsilon|x - x_0| \)越小，进而\( f(x) \)和\( f(x_0) + L(x - x_0) \)越接近。同理，当\( x \)和\( x_0 \)越接近时，\( |x - x_0| \)越小，从而\( \epsilon|x - x_0| \)越小，进而\( f(x) \)和\( f(x_0) + L(x - x_0) \)越接近。结论就是，当\( \epsilon \)越小、\( x \)和\( x_0 \)越接近时， \( f(x) \)和\( f(x_0) + L(x - x_0) \)就越接近，此时可以用\( f(x_0) + L(x - x_0) = f(x_0) + f'(x_0)(x - x_0) \) 来近似\( f(x) \)的值。当然，\( \epsilon \)取较小时，区间范围\( \delta \) 也会变小，此时可近似的\( x \)的范围\( x \in X, |x - x_0| \leq \delta \)也会变小。

证明：

\( 1 \Longrightarrow 2 \)：

如果1成立，即\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = L \)，则\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta, |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)，可得\( |x - x_0||\dfrac{f(x) - f(x_0)}{x - x_0} - L| = |(x - x_0)(\dfrac{f(x) - f(x_0)}{x - x_0} - L)| = |f(x) - f(x_0) - L(x - x_0)| = |f(x) - (f(x_0) + L(x - x_0))| \leq \epsilon|x - x_0| \)。而针对\( x = x_0 \)，有\( |x - x_0| = 0 \leq \delta \)，以及\( |f(x) - (f(x_0) + L(x - x_0))| = |f(x_0) - (f(x_0) + L(x_0 - x_0))| = 0 \leq \epsilon \)。综上，\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X, |x - x_0| \leq \delta, |f(x) - (f(x_0) + L(x - x_0))| \leq \epsilon|x - x_0| \)，即2成立。

\( 2 \Longrightarrow 1 \)：

如果2成立，即\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X, |x - x_0| \leq \delta, |f(x) - (f(x_0) + L(x - x_0))| \leq \epsilon|x - x_0| \)，可得\( \dfrac{|f(x) - (f(x_0) + L(x - x_0))|}{|x - x_0|} = \dfrac{|f(x) - f(x_0) - L(x - x_0)|}{|x - x_0|} = |\dfrac{f(x) - f(x_0)}{x - x_0} - \dfrac{L(x - x_0)}{x - x_0}| = |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)，而\( \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta \)，有\( x \in X, |x - x_0| \leq \delta \)，进而有\( |\dfrac{f(x) - f(x_0)}{x - x_0} - L| \leq \epsilon \)，可得\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = L \)，即1成立。

证毕。

练习10.1.3

题目：

Prove Proposition 10.1.10. (Hint: either use the limit laws (Proposition 9.3.14), or use Proposition 10.1.7.)

Proposition 10.1.10的内容：

(Differentiability implies continuity). Let \( X \) be a subset of \( \mathbf{R} \), let \( x_0 \in X \) be a limit point of \( X \), and let \( f: X \to \mathbf{R} \) be a function. If \( f \) is differentiable at \( x_0 \), then \( f \) is also continuous at \( x_0 \).

证明：

因为\( f \)可微，因此\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \) 存在，又\( \lim_{x \to x_0; x \in X - \{ x_0 \}} (x - x_0) = 0 \)，根据定理9.3.14，可得\( \lim_{x \to x_0; x \in X - \{ x_0 \}} (f(x) - f(x_0)) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} \times \lim_{x \to x_0; x \in X - \{ x_0 \}} (x - x_0) = 0 \)，再次根据定理9.3.14，可得\( \lim_{x \to x_0; x \in X - \{ x_0 \}} f(x) = \lim_{x \to x_0; x \in X - \{ x_0 \}} (f(x) - f(x_0)) + \lim_{x \to x_0; x \in X - \{ x_0 \}} f(x_0) = 0 + f(x_0) = f(x_0) \)，于是有 \( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta, |f(x) - f(x_0)| \leq \epsilon \)，而针对\( x = x_0 \)，有\( |x - x_0| = 0 \leq \delta \)以及 \( |f(x) - f(x_0)| = |f(x_0) - f(x_0)| = 0 \leq \epsilon \)。综上，有\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X, |x - x_0| \leq \delta, |f(x) - f(x_0)| \leq \epsilon \)，即\( f \)在\( x_0 \)连续。

证毕。

练习10.1.4

题目：

Prove Theorem 10.1.13. (Hint: use the limit laws in Proposition 9.3.14. Use earlier parts of this theorem to prove the latter. For the product rule, use the identity \( f(x)g(x) - f(x_0)g(x_0) = f(x)g(x) - f(x)g(x_0) + f(x)g(x_0) - f(x_0)g(x_0) = f(x)(g(x) - g(x_0)) + (f(x) - f(x_0))g(x_0) \); this trick of adding and subtracting an intermediate term is sometimes known as the “middle-man trick” and is very useful in analysis.)

Theorem 10.1.13的内容：

(Differential calculus). Let \( X \) be a subset of \( \mathbf{R} \), let \( x_0 \in X \) be a limit point of \( X \), and let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions.

1. If \( f \) is a constant function, i.e., there exists a real number \( c \) such that \( f(x) = c \) for all \( x \in X \), then \( f \) is differentiable at \( x_0 \) and \( f'(x_0) = 0 \).
2. If \( f \) is the identity function, i.e., \( f(x) = x \) for all \( x \in X \), then \( f \) is differentiable at \( x_0 \) and \( f'(x_0) = 1 \).
3. (Sum rule) If \( f \) and \( g \) are differentiable at \( x_0 \), then \( f + g \) is also differentiable at \( x_0 \), and \( (f + g)'(x_0) = f'(x_0) + g'(x_0) \).
4. (Product rule) If \( f \) and \( g \) are differentiable at \( x_0 \), then \( fg \) is also differentiable at \( x_0 \), and \( (fg)'(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0) \).
5. If \( f \) is differentiable at \( x_0 \) and \( c \) is real number, then \( cf \) is also differentiable at \( x_0 \), and \( (cf)'(x_0) = cf'(x_0) \).
6. (Difference rule) If \( f \) and \( g \) are differentiable at \( x_0 \), then \( f - g \) is also differentiable at \( x_0 \), and \( (f - g)'(x 0 ) = f'(x_0) - g'(x_0) \).
7. If \( g \) is differentiable at \( x_0 \), and \( g \) is non-zero on \( X \) (i.e., \( g(x) \neq 0 \) for all \( x \in X \)), then \( 1/g \) is also differentiable at \( x_0 \), and \( (\dfrac{1}{g})'(x_0) = -\dfrac{g'(x_0)}{g(x_0)^2} \).
8. (Quotient rule) If \( f \) and \( g \) are differentiable at \( x_0 \), and \( g \) is non-zero on \( X \), then \( f/g \) is also differentiable at \( x_0 \), and \( (\dfrac{f}{g})'(x_0) = -\dfrac{f'(x_0)g(x_0) - f(x_0)g'(x_0)}{g(x_0)^2} \).

证明1：

根据定理9.3.14，可得 \( f'(x_0) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{c - c}{x - x_0} = \dfrac{\lim_{x \to x_0; x \in X - \{ x_0 \}} {(c - c)}}{\lim_{x \to x_0; x \in X - \{ x_0 \}} (x - x_0)} = 0 \)。

证毕。

证明2：

根据定理9.3.14，可得 \( f'(x_0) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{x - x_0}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} 1 = 1 \)。

证毕。

证明3：

根据定理9.3.14以及\( f, g \)均在\( x_0 \)可微，可得 \( (f + g)'(x_0) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{(f + g)(x) - (f + g)(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) + g(x) - f(x_0) - g(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{(f(x) - f(x_0)) + (g(x) - g(x_0))}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} + \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{g(x) - g(x_0)}{x - x_0} = f'(x_0) + g'(x_0) \)。

证毕。

证明4：

根据\( f, g \)均在\( x_0 \)可微以及推论10.1.12，可得\( \lim_{x \to x_0; x \in X - \{ x_0 \}} f(x) = f(x_0), \lim_{x \to x_0; x \in X - \{ x_0 \}} g(x) = g(x_0) \)。接着，根据定理9.3.14以及\( f, g \)均在\( x_0 \)可微，可得 \( (fg)'(x_0) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{(fg)(x) - (fg)(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x)g(x) - f(x_0)g(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x)g(x) - f(x)g(x_0) + f(x)g(x_0) - f(x_0)g(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x)(g(x) - g(x_0)) + (f(x) - f(x_0))g(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x)(g(x) - g(x_0))}{x - x_0} + \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{(f(x) - f(x_0))g(x_0)}{x - x_0} = ((\lim_{x \to x_0; x \in X - \{ x_0 \}} f(x))(\lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{g(x) - g(x_0)}{x - x_0})) + ((\lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0})(\lim_{x \to x_0; x \in X - \{ x_0 \}} g(x_0))) = f(x_0)g'(x_0) + f'(x_0)g(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0) \)。

证毕。

证明5：

定义函数\( h(x): X \to \mathbf{R}, \forall x \in X \)，令\( h(x) := c \)，根据1，有\( h(x) \)在\( x_0 \)可微且\( h'(x_0) = 0 \)，接着根据4，有\( cf = hf \)可微，且\( (cf)'(x_0) = (hf)'(x_0) = h'(x_0)f(x_0) + h(x_0)f'(x_0) = 0 + cf'(x_0) = cf'(x_0) \)。

证毕。

证明6：

根据5，有\( -g \)在\( x_0 \)可微且\( (-g)'(x_0) = -g'(x_0) \)，接着根据3，有\( f - g = f + (-g) \)在\( x_0 \)可微且\( (f - g)'(x_0) = (f + (-g))'(x_0) = f'(x_0) + (-g)'(x_0) = f'(x_0) - g'(x_0) \)。

证毕。

证明7：

根据\( g \)均在\( x_0 \)可微以及推论10.1.12，可得\( \lim_{x \to x_0; x \in X - \{ x_0 \}} g(x) = g(x_0) \)。

根据定理9.3.14、\( \forall x \in X, g(x) \neq 0 \)以及\( g \)均在\( x_0 \)可微，可得 \( (\dfrac{1}{g})'(x_0) = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{(\dfrac{1}{g})(x) - (\dfrac{1}{g})(x_0)}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{\dfrac{1}{g(x)} - \dfrac{1}{g(x_0)}}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{\dfrac{g(x_0) - g(x)}{g(x)g(x_0)}}{x - x_0} = \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{g(x_0) - g(x)}{(x - x_0)(g(x)g(x_0))} = (\lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{g(x_0) - g(x)}{x - x_0}) (\lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{1}{g(x)g(x_0)}) = -g'(x_0)(\dfrac{\lim_{x \to x_0; x \in X - \{ x_0 \}} 1}{\lim_{x \to x_0; x \in X - \{ x_0 \}} g(x)g(x_0)}) = -g'(x_0)(\dfrac{1}{(\lim_{x \to x_0; x \in X - \{ x_0 \}} g(x))(\lim_{x \to x_0; x \in X - \{ x_0 \}} g(x_0))}) = -g'(x_0)(\dfrac{1}{g(x_0)g(x_0)}) = -\dfrac{g'(x_0)}{g(x_0)^2} \)。

证毕。

证明8：

由\( g \)在\( x_0 \)可微、\( \forall x \in X, g(x) \neq 0 \)以及7，可得\( \dfrac{1}{g} \)在\( x_0 \)可微，且\( (\dfrac{1}{g})'(x_0) = -\dfrac{g'(x_0)}{g(x_0)^2} \)。接着由4可得，\( (\dfrac{f}{g}) = f(\dfrac{1}{g}) \)在\( x_0 \)可微且 \( (\dfrac{f}{g})'(x_0) = (f(\dfrac{1}{g}))'(x_0) = f'(x_0)(\dfrac{1}{g})(x_0) + f(x_0)(\dfrac{1}{g})'(x_0) = \dfrac{f'(x_0)}{g(x_0)} + f(x_0) (-\dfrac{g'(x_0)}{g(x_0)^2}) = \dfrac{f'(x_0)}{g(x_0)} + \dfrac{-f(x_0)g'(x_0)}{g(x_0)^2} = \dfrac{f'(x_0)g(x_0)}{g(x_0)^2} + \dfrac{-f(x_0)g'(x_0)}{g(x_0)^2} = \dfrac{f'(x_0)g(x_0) - f(x_0)g'(x_0)}{g(x_0)^2} \)。

证毕。

练习10.1.5

题目：

Let \( n \) be a natural number, and let \( f: \mathbf{R} \to \mathbf{R} \) be the function \( f(x) := x^n \). Show that \( f \) is differentiable on \( \mathbf{R} \) and \( f'(x) = nx^{n - 1} \) for all \( x \in \mathbf{R} \) with the convention that \( nx^{n - 1} = 0 \) when \( n = 0 \). (Hint: use Theorem 10.1.13 and induction.)

注：

“with the convention that \( nx^{n - 1} = 0 \) when \( n = 0 \)“这句话是作者在博客勘误中加的，实际上可不加，因为我们只说了\( n \)是自然数，并没有说\( n - 1 \)是自然数。

证明：

使用数学归纳法证明：

当\( n = 0 \)时，\( \forall x_0 \in \mathbf{R}, f(x_0) := (x_0)^0 = 1 \)，此时根据定理10.1.13的1，有\( f'(x_0) = 0 = n(x_0)^{n - 1} \)，基础情况成立。

归纳假设当\( n = k \)时成立，当\( n = k + 1 \)时，定义函数\( g: \mathbf{R} \to \mathbf{R}, \forall x \in \mathbf{R} \)，令\( g(x) := x^k \)，定义函数\( h: \mathbf{R} \to \mathbf{R}, \forall x \in \mathbf{R} \)，令\( h(x) := x \)，可得\( f = gh \)。\( \forall x_0 \in \mathbf{R} \)，根据归纳假设，有\( g \)在\( x_0 \)可微且\( g'(x_0) = k(x_0)^{k - 1} \)，根据定理10.1.13的4以及\( h \)在\( x_0 \)可微，有\( f = gh \)在\( x_0 \)可微且\( f'(x_0) = (gh)'(x_0) = g'(x_0)h(x_0) + g(x_0)h'(x_0) = k(x_0)^{k - 1}x_0 + (x_0)^k \times 1 = k(x_0)^k + (x_0)^k = (k + 1)(x_0)^k \)，即\( n = k + 1 \)时也成立，归纳完毕。

证毕。

练习10.1.6

题目：

Let \( n \) be a negative integer, and let \( f : \mathbf{R} - \{ 0 \} \to \mathbf{R} \) be the function \( f(x) := x^n \). Show that \( f \) is differentiable on \( \mathbf{R} - \{ 0 \} \) and \( f'(x) = nx^{n - 1} \) for all \( x \in \mathbf{R} - \{ 0 \} \). (Hint: use Theorem 10.1.13 and Exercise 10.1.5.)

证明：

令\( m := -n \)，有\( m \)为正自然数且\( \forall x \in \mathbf{R} - \{ 0 \}, f(x) := (x)^n = (x)^{-m} = \dfrac{1}{(x)^m} \)。定义函数\( g: \mathbf{R} - \{ 0 \} \to \mathbf{R}, \forall x \in \mathbf{R} - \{ 0 \} \)，令\( g(x) := x^m \)，可得\( f = \dfrac{1}{g} \)。 \( \forall x_0 \in \mathbf{R} - \{ 0 \} \)，有\( g(x_0) \neq 0 \)，根据练习10.1.5，可得\( g'(x_0) = mx^{m -1} \)，再定理10.1.13的7，可得\( f'(x_0) = -\dfrac{g'(x_0)}{g(x_0)^2} = -\dfrac{m(x_0)^{m - 1}}{(x_0)^{2m}} = -m(x_0)^{m - 1 - 2m} = -m(x_0)^{-m - 1} = n(x_0)^{n - 1} \)。

证毕。

练习10.1.7

题目：

Prove Theorem 10.1.15. (Hint: one way to do this is via Newton’s approximation, Proposition 10.1.7. Another way is to use Proposition 9.3.9 and Proposition 10.1.10 to convert this problem into one involving limits of sequences, however with the latter strategy one has to treat the case \( f'(x_0) = 0 \) separately, as some division-by-zero subtleties can occur in that case.)

Theorem 10.1.15的内容：

(Chain rule). Let \( X, Y \) be subsets of \( \mathbf{R} \), let \( x_0 \in X \) be a limit point of \( X \), and let \( y_0 \in Y \) be a limit point of \( Y \). Let \( f: X \to Y \) be a function such that \( f(x_0) = y_0 \), and such that \( f \) is differentiable at \( x_0 \). Suppose that \( g: Y \to \mathbf{R} \) is a function which is differentiable at \( y_0 \). Then the function \( g \circ f: X \to \mathbf{R} \) is differentiable at \( x_0 \), and \( (g \circ f)'(x_0) = g'(y_0)f'(x_0) \).

注：

这里不按Hint的思路来，直接用可微的定义来证，不过需要定义辅助函数来处理 \( f(x) - f(x_0) = 0 \)导致\( \dfrac{g(f(x)) - g(f(x_0))}{f(x) - f(x_0)} \)未定义的情况。

证明：

由\( f \)在\( X \)上的\( x_0 \)可微，可得 \( \lim_{x \to x_0; X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = f'(x_0) \)。

定义函数\( d: X \to Y, \forall x \in X \)，如果\( f(x) - f(x_0) = 0 \)，则令\( d(x) := 0 \)，否则，令\( d(x) := \dfrac{g(f(x)) - g(f(x_0))}{f(x) - f(x_0)} - g'(f(x_0)) \)， \( \forall x \in X \)，易得不管是\( f(x) - f(x_0) = 0 \)，还是\( f(x) - f(x_0) \neq 0 \)，均有\( g(f(x)) - g(f(x_0)) = (d(x) + g'(f(x_0)))(f(x) - f(x_0)) \) （分情况讨论下即可得到该等式，该等式意味着等下证极限的时候，可以将\( g(f(x)) - g(f(x_0)) \)改写成\( (d(x) + g'(f(x_0)))(f(x) - f(x_0)) \)）。

我们证明下\( \lim_{x \to x_0; X - \{ x_0 \}} d(x) = 0 \)：由\( g \)在\( X \)上的\( f(x_0) \)可微，可得 \( \lim_{y \to f(x_0); Y - \{ f(x_0) \}} \dfrac{g(y) - g(f(x_0))}{y - f(x_0)} = g'(f(x_0)) \)，于是有\( \forall \epsilon > 0, \exists \delta_1 > 0, \forall y \in Y - \{ f(x_0) \}, |y - f(x_0)| \leq \delta_1, |\dfrac{g(y) - g(f(x_0))}{y - f(x_0)} - g'(f(x_0))| \leq \epsilon \) (1) ，由\( f \)在\( X \)上的\( x_0 \)可微以及定理10.1.10，可得\( f \)在\( X \)上的\( x_0 \)连续，于是有\( \lim_{x \to x_0; x \in X} f(x) = f(x_0) \)，进而有\( \exists \delta > 0, \forall x \in X, |x - x_0| \leq \delta, |f(x) - f(x_0)| \leq \delta_1 \) (2) ，针对\( \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta \)，有\( x \in X, |x - x_0| \leq \delta \)，于是根据(2)，也有\( |f(x) - f(x_0)| \leq \delta_1 \)，接下来分情况讨论：

1. 如果\( f(x) - f(x_0) \neq 0 \)，此时有\( f(x) \in Y - \{ f(x_0) \} \)，根据(1)，可得\( |\dfrac{g(f(x)) - g(f(x_0))}{f(x) - f(x_0)} - g'(f(x_0))| \leq \epsilon \)，即\( |d(x) - 0| \leq \epsilon \)。
2. 如果\( f(x) - f(x_0) = 0 \)，此时有\( d(x) = 0 \)，进而有\( |d(x) - 0| = |0 - 0| \leq \epsilon \)。

综上，有\( \forall \epsilon > 0, \exists \delta > 0, \forall x \in X - \{ x_0 \}, |x - x_0| \leq \delta, |d(x) - 0| \leq \epsilon \)，可得\( \lim_{x \to x_0; X - \{ x_0 \}} d(x) = 0 \)。

现在可以证明命题的结论了，\( \lim_{x \to x_0; X - \{ x_0 \}} \dfrac{(g \circ f)(x) - (g \circ f)(x_0)}{x - x_0} = \lim_{x \to x_0; X - \{ x_0 \}} \dfrac{g(f(x)) - g(f(x_0))}{x - x_0} = \lim_{x \to x_0; X - \{ x_0 \}} \dfrac{(d(x) + g'(f(x_0)))(f(x) - f(x_0))}{x - x_0} = (\lim_{x \to x_0; X - \{ x_0 \}} (d(x) + g'(f(x_0)))) (\lim_{x \to x_0; X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0}) = (\lim_{x \to x_0; X - \{ x_0 \}} d(x) + \lim_{x \to x_0; X - \{ x_0 \}} g'(f(x_0)))f'(x_0) = (0 + g'(f(x_0)))f'(x_0) = g'(f(x_0))f'(x_0) \)，这意味着\( g \circ f \)在\( X \)上的\( x_0 \)可微且\( (g \circ f)'(x_0) = g'(f(x_0))f'(x_0) = g'(y_0)f'(x_0) \)。

证毕。

章节10.2

练习10.2.1

题目：

Prove Proposition 10.2.6.

Proposition 10.2.6的内容：

(Local extrema are stationary). Let \( a < b \) be real numbers, and let \( f: (a, b) \to \mathbf{R} \) be a function. If \( x_0 \in (a, b) \), \( f \) is differentiable at \( x_0 \), and \( f \) attains either a local maximum or local minimum at \( x_0 \), then \( f'(x_0) = 0 \).

证明：

不妨设\( f \)在\( x_0 \)达到局部最大值（达到局部最小值的情况类似，这里省略），可得\( \exists \delta_1 > 0, \forall x \in (a, b) \cap (x_0 - \delta_1 , x_0 + \delta_1), f(x) \leq f(x_0) \) (1) ，假设\( f'(x_0) \neq 0 \)，即\( f'(x_0) > 0 \)或\( f'(x_0) < 0 \)，不妨设\( f'(x_0) > 0 \)（\( f'(x_0) < 0 \)的情况类似，这里省略），根据定理10.1.7，可得\( \exists \delta_2 > 0, \forall x \in (a, b), |x - x_0| \leq \delta_2, |f(x) - f(x_0) - f'(x_0)(x - x_0)| \leq \dfrac{f'(x_0)}{2}|x - x_0| \)，令\( \delta := \min(\delta_1, \delta_2) \)，可得\( \forall x \in (a, b), |x - x_0| \leq \delta, |f(x) - f(x_0) - f'(x_0)(x - x_0)| \leq \dfrac{f'(x_0)}{2}|x - x_0| \) (2) ，令\( x_1 := x_0 + \min(\dfrac{b - x_0}{2}, \dfrac{\delta}{2}) \) （注：如果是\( f'(x_0) < 0 \)的情况，就换成\( x_1 := x_0 - \min(\dfrac{x_0 - a}{2}, \dfrac{\delta}{2}) \)），有\( x_1 \in (a, b), x_1 \in (a, b) \cap (x_0 - \delta_1 , x_0 + \delta_1), x_1 > x_0, |x_1 - x_0| \leq \delta \)，根据(2)，有 \( |f(x_1) - f(x_0) - f'(x_0)(x_1 - x_0)| \leq \dfrac{f'(x_0)}{2}|x_1 - x_0| \)，又\( x_1 > x_0 \)，因此\( |x_1 - x_0| = x_1 - x_0 \)，可得\( -\dfrac{f'(x_0)}{2}(x_1 - x_0) \leq f(x_1) - f(x_0) - f'(x_0)(x_1 - x_0) \leq \dfrac{f'(x_0)}{2}(x_1 - x_0) \)，进一步可得\( (-\dfrac{f'(x_0)}{2} + f'(x_0))(x_1 - x_0) \leq f(x_1) - f(x_0) \leq (\dfrac{f'(x_0)}{2} + f'(x_0))(x_1 - x_0) \)，因为\( f'(x_0) > 0 \)，因此\( (-\dfrac{f'(x_0)}{2} + f'(x_0))(x_1 - x_0) > 0, (\dfrac{f'(x_0)}{2} + f'(x_0))(x_1 - x_0) > 0 \)，可得\( f(x_1) - f(x_0) > 0 \)，即\( f(x_1) > f(x_0) \)，但\( x_1 \in (a, b) \cap (x_0 - \delta_1 , x_0 + \delta_1) \)，这和(1)矛盾，因此假设不成立，有\( f'(x_0) = 0 \)。

证毕。

练习10.2.2

题目：

Give an example of a function \( f: (−1, 1) \to \mathbf{R} \) which is continuous and attains a global maximum at \( 0 \), but which is not differentiable at \( 0 \). Explain why this does not contradict Proposition 10.2.6.

例子：

定义函数\( f: (-1, 1) \to \mathbf{R}, \forall x \in (-1, 1) \)，如果\( x \leq 0 \)，则令\( f(x) := x \)，如果\( x > 0 \)，则令\( f(x) := -x \)，易证\( f(x) \)连续，同时有\( \forall x \in (-1, 1), f(x) \leq 0 \)，即\( f(x) \)在\( 0 \)处达到全局最大值，但由于\( \lim_{x \to 0; x \in (-1, 0)} \dfrac{f(x) - f(0)}{x - 0} = \lim_{x \to 0; x \in (-1, 0)} \dfrac{x - 0}{x - 0} = 1 \)，而\( \lim_{x \to 0; x \in (0, 1)} \dfrac{f(x) - f(0)}{x - 0} = \lim_{x \to 0; x \in (0, 1)} \dfrac{-x - 0}{x - 0} = -1 \neq \lim_{x \to 0; x \in (-1, 0)} \dfrac{f(x) - f(0)}{x - 0} \)，因此\( \lim_{x \to 0; x \in (-1, 1) - \{ 0 \}} \dfrac{f(x) - f(0)}{x - 0} \)不存在，\( f(x) \)在\( 0 \)处不可微。

解释上面例子为什么没有和定理10.2.6矛盾：

使用定理10.2.6的前提条件之一是\( f \)在\( x_0 \)处可微（上面例子中\( x_0 = 0 \)），而上面例子不满足该前提条件，无法使用定理10.2.6，故不矛盾。

练习10.2.3

题目：

Give an example of a function \( f: (−1, 1) \to \mathbf{R} \) which is differentiable, and whose derivative equals \( 0 \) at \( 0 \), but such that \( 0 \) is neither a local minimum nor a local maximum. Explain why this does not contradict Proposition 10.2.6.

例子：

定义函数\( f: (-1, 1) \to \mathbf{R}, \forall x \in (-1, 1) \)，令\( f(x) := x^3 \)，根据练习10.1.1、练习10.1.5，有\( f'(0) = 3 \times 0^2 = 0 \)，但\( \forall \delta > 0 \)，有\( 0 + \min(\dfrac{\delta}{2}, \dfrac{1}{2}) \in (-1, 1) \cap (0 - \delta, 0 + \delta), 0 - \min(\dfrac{\delta}{2}, \dfrac{1}{2}) \in (-1, 1) \cap (0 - \delta, 0 + \delta) \) 但\( f(0 + \min(\dfrac{\delta}{2}, \dfrac{1}{2})) > f(0), f(0 - \min(\dfrac{\delta}{2}, \dfrac{1}{2})) < f(0) \)，因此\( f \)在\( 0 \)处即没有达到局部最大值，也没有达到局部最小值。

解释上面例子为什么没有和定理10.2.6矛盾：

简单来讲就是上面例子跟定理10.2.6是反着来的，跟定理10.2.6的正确与否没什么关系，更严格点讲就是使用定理10.2.6的前提条件之一是\( f \)在\( x_0 \)处达到局部最大值或者局部最小值（上面例子中\( x_0 = 0 \)），而上面例子不满足该前提条件，无法使用定理10.2.6，故不矛盾。

练习10.2.4

题目：

Prove Theorem 10.2.7. (Hint: use the maximum principle, Proposition 9.6.7, followed by Proposition 10.2.6. Note that the maximum principle does not tell you whether the maximum or minimum is in the open interval \( (a, b) \) or is one of the boundary points \( a \), \( b \), so you have to divide into cases and use the hypothesis \( g(a) = g(b) \) somehow.)

Theorem 10.2.7的内容：

(Rolle’s theorem). Let \( a < b \) be real numbers, and let \( g: [a, b] \to \mathbf{R} \) be a continuous function which is differentiable on \( (a, b) \). Suppose also that \( g(a) = g(b) \). Then there exists an \( x \in (a, b) \) such that \( g'(x) = 0 \).

证明：

由\( g \)在\( [a, b] \)上连续以及定理9.6.7，可得\( \exists x_{\max} \in [a, b], x_{\min} \in [a, b] \)， \( g \)在\( x_{\max} \)处达到最大值，在\( x_{\min} \)处达到最小值。

如果\( x_{\max}, x_{\min} \)均\( \notin (a, b) \)，即\( x_{\max}, x_{\min} \)均是端点，分情况讨论：

1. 如果\( x_{\max} = a, x_{\min} = a \)，此时\( \forall x \in [a, b], g(a) \leq g(x) \leq g(a) \)，可得\( g(x) = g(a) \)。
2. 如果\( x_{\max} = b, x_{\min} = b \)，此时\( \forall x \in [a, b], g(b) \leq g(x) \leq g(b) \)，可得\( g(x) = g(b) \)。
3. 如果\( x_{\max} = a, x_{\min} = b \)，此时\( \forall x \in [a, b], g(b) \leq g(x) \leq g(a) = g(b) \)，可得\( g(x) = g(b) \)。
4. 如果\( x_{\max} = b, x_{\min} = a \)，此时\( \forall x \in [a, b], g(a) \leq g(x) \leq g(b) = g(a) \)，可得\( g(x) = g(a) \)。

综上，所有情况下，\( f \)都是常数函数，因此根据定理10.1.13的1，有\( \forall x \in [a, b], g'(x) = 0 \)，特别的，有\( \dfrac{a + b}{2} \in (a, b), g'(\dfrac{a + b}{2}) = 0 \)，符合命题的结论。

如果\( x_{\max}, x_{\min} \)中其中一个\( \in (a, b) \)，不妨设\( x_{\max} \in (a, b) \)，则根据\( f \)在\( (a, b) \)可微以及根据定理10.2.6，可得\( f'(x_{\max}) = 0 \)。

证毕。

练习10.2.5

题目：

Use Theorem 10.2.7 to prove Corollary 10.2.9. (Hint: consider a function of the form \( f(x) - cx \) for some carefully chosen real number \( c \).)

Theorem 10.2.9的内容：

(Mean value theorem). Let \( a < b \) be real numbers, and let \( f: [a, b] \to \mathbf{R} \) be a function which is continuous on \( [a, b] \) and differentiable on \( (a, b) \). Then there exists an \( x \in (a, b) \) such that \( f'(x) = \dfrac{f(b) - f(a)}{b - a} \).

证明：

令\( c := \dfrac{f(b) - f(a)}{b - a} \)，定义函数\( g: [a, b] \to \mathbf{R}, \forall x \in [a, b] \)，令\( g(x) := f(x) - cx \)，由函数\( cx \)在\( [a, b] \)上连续以及定理9.4.9，可得\( g \)也在\( [a, b] \)上连续，由函数\( cx \)在\( (a, b) \)上可微以及定理10.1.13的6，可得\( g \)也在\( (a, b) \)上可微。而\( g(a) = f(a) - ca = \dfrac{f(a)b - f(a)a - f(b)a + f(a)a}{b - a} = \dfrac{f(a)b - f(b)a}{b - a}, g(a) = f(b) - cb = \dfrac{f(b)b - f(b)a - f(b)b + f(a)b}{b - a} = \dfrac{f(a)b - f(b)a}{b - a} \)，有\( g(a) = g(b) \)，根据定理10.2.7，可得 \( \exists x \in (a, b), g'(x) = 0 \)，再根据定理10.1.13的6（或者3），有\( f'(x) = g'(x) + c = 0 + c = \dfrac{f(b) - f(a)}{b - a} \)。

证毕。

练习10.2.6

题目：

Let \( M > 0 \), and let \( f: [a, b] \to \mathbf{R} \) be a function which is continuous on \( [a, b] \) and differentiable on \( (a, b) \), and such that \( |f'(x)| \leq M \) for all \( x \in (a, b) \) (i.e., the derivative of \( f \) is bounded). Show that for any \( x, y \in [a, b] \) we have the inequality \( |f(x) - f(y)| \leq M |x - y| \). (Hint: apply the mean value theorem (Corollary 10.2.9) to a suitable restriction of \( f \).) Functions which obey the bound \( |f(x) - f(y)| \leq M|x - y| \) are known as Lipschitz continuous functions with Lipschitz constant \( M \); thus this exercise shows that functions with bounded derivative are Lipschitz continuous.

证明：

\( \forall x, y \in [a, b] \)，如果\( x = y \)，则有\( |f(x) - f(y)| = 0 \leq M|x - y| \)，如果\( x \neq y \)，则有\( x < y \)或\( y < x \)，不妨设\( x < y \)，此时，因为\( [x, y] \subseteq [a, b] \)且\( f \)在\( [a, b] \)上连续，根据练习9.4.6，有\( f|_{[x, y]} \)在\( [x, y] \)上连续，又因为\( [x, y] \subseteq [a, b] \)且\( f \)在\( (a, b) \)上可微，根据练习10.1.1，可得\( f|_{[x, y]} \)在\( (x, y) \)上可微（注：针对\( (x, y) \)内的每一点使用练习10.1.1即可），综上，有\( f|_{[x, y]} \)在\( [x, y] \)上连续，在\( (x, y) \)上可微，根据定理10.2.9，\( \exists x_0 \in (x, y), (f|_{[x, y]})'(x_0) = \dfrac{f|_{[x, y]}(y) - f|_{[x, y]}(x)}{y - x} \)，又根据练习10.1.1，有\( |(f|_{[x, y]})'(x_0)| = |f'(x_0)| \leq M \)，因此有\( |\dfrac{f|_{[x, y]}(y) - f|_{[x, y]}(x)}{y - x}| = \dfrac{|f|_{[x, y]}(y) - f|_{[x, y]}(x)|}{|y - x|} \leq M \)，可得\( |f|_{[x, y]}(x) - f|_{[x, y]}(y)| = |f|_{[x, y]}(y) - f|_{[x, y]}(x)| \leq M|y - x| = M|x - y| \)，进而可得\( |f(x) - f(y)| = |f|_{[x, y]}(x) - f|_{[x, y]}(y)| \leq M|x - y| \)。

综上，\( \forall x, y \in [a, b] \)，均有\( |f(x) - f(y)| \leq M|x - y| \)。

证毕。

练习10.2.7

题目：

Let \( f: \mathbf{R} \to \mathbf{R} \) be a differentiable function such that \( f' \) is bounded. Show that \( f \) is uniformly continuous. (Hint: use the preceding exercise.)

证明：

因为\( f \)在\( \mathbf{R} \)上可微，因此根据定理10.1.10，有\( f \)在\( \mathbf{R} \)上连续。

因为\( f' \)有界，因此\( \exists M > 0, \forall x \in \mathbf{R}, |f'(x)| \leq M \) (1) 。

\( \forall \epsilon > 0 \)，令\( \delta := \dfrac{\epsilon}{M} \)，此时\( \forall x, y \in \mathbf{R}, |x - y| \leq \delta \)，如果\( x = y \)，则\( |f(x) - f(y)| = 0 \leq \epsilon \)，如果\( x \neq y \)，则有\( x < y \)或\( y < x \)，不妨设\( x < y \)，此时根据(1)，有\( \forall z \in (x, y), |f'(z)| \leq M \) (2) ，因为\( [x, y] \subseteq \mathbf{R} \)且\( f \)在\( \mathbf{R} \)上连续，因此根据练习9.4.6，有\( f|_{[x, y]} \)在\( [x, y] \)上连续，因为\( [x, y] \subseteq \mathbf{R} \)且\( f \)在\( \mathbf{R} \)上可微，根据练习10.1.1，可得\( f|_{[x, y]} \)在\( (x, y) \)上可微（注：针对\( (x, y) \)内的每一点使用练习10.1.1即可），综上，有\( f|_{[x, y]} \)在\( [x, y] \)上连续，在\( (x, y) \)上可微，根据(2)和练习10.2.6，有\( \forall z, w \in [x, y], |f(z) - f(w)| \leq M|z - w| \)，而\( |z - w| \leq |x - y| \leq \delta \)，因此\( |f(z) - f(w)| \leq M|z - w| \leq M \delta = M \dfrac{\epsilon}{M} = \epsilon \)，特别的，\( x, y \in [x, y] \)，因此有\( |f(x) - f(y)| \leq \epsilon \)。综上，\( \forall \epsilon > 0, \exists \delta > 0, \forall x, y \in \mathbf{R}, |x - y| \leq \epsilon \)，可得\( f \)一致连续。

证毕。

章节10.3

练习10.3.1

题目：

Prove Proposition 10.3.1.

Proposition 10.3.1的内容：

Let \( X \) be a subset of \( \mathbf{R} \), let \( x_0 \in X \) be a limit point of \( X \), and let \( f: X \to \mathbf{R} \) be a function. If \( f \) is monotone increasing and \( f \) is differentiable at \( x_0 \), then \( f'(x_0) \geq 0 \). If \( f \) is monotone decreasing and \( f \) is differentiable at \( x_0 \), then \( f'(x_0) \leq 0 \).

证明：

如果\( f \)单调递增且\( f \)在\( x_0 \)可微，则有\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = f'(x_0) \)，定义函数\( g(x) := \dfrac{f(x) - f(x_0)}{x - x_0} \)，则根据定理9.3.9，有\( \forall \)序列\( (a_n)_{n = 0}^{\infty} \)满足 \( \forall n \in \mathbf{N}, a_n \in X - \{ x_0 \} \)且\( \lim_{n \to \infty} a_n = x_0 \)，有\( \lim_{n \to \infty} g(a_n) = f'(x_0) \) (1) ，又\( x_0 \)是\( X \)的极限点，进而是\( X - \{ x_0 \} \)的聚点，于是根据引理9.1.14，存在序列\( (b_n)_{n = 0}^{\infty} \)满足 \( \forall n \in \mathbf{N}, b_n \in X - \{ x_0 \} \)且\( \lim_{n \to \infty} b_n = x_0 \)，根据(1)，可得\( \lim_{n \to \infty} g(b_n) = f'(x_0) \)。针对\( \forall n \in \mathbf{N} \)，如果\( b_n > x_0 \)，则\( b_n - x_0 > 0 \)，又\( f \)单调递增，可得\( f(b_n) - f(x_0) \geq 0 \)，于是有\( g(b_n) = \dfrac{f(b_n) - f(x_0)}{b_n - x_0} \geq 0 \)，如果\( b_n < x_0 \)，则\( b_n - x_0 < 0 \)，又\( f \)单调递增，可得\( f(b_n) - f(x_0) \leq 0 \)，于是还是有\( g(b_n) = \dfrac{f(b_n) - f(x_0)}{b_n - x_0} \geq 0 \)，又\( b_n \in X - \{ x_0 \} \)，故不用考虑\( b_n = x_0 \)的情况，综上可得，\( \forall n \in \mathbf{N}, g(b_n) \geq 0 \)。对序列\( (g(b_n))_{n = 0}^{\infty}, (0)_{n = 0}^{\infty} \)使用引理6.4.13 （或者直接用定理5.4.9的实数版本，不过本书没证这个），可得\( \lim \sup_{n \to \infty} g(b_n) \geq \lim \sup_{n \to \infty} 0 = 0 \)，再由(1)以及定理6.4.12的6，可得\( \lim_{n \to \infty} g(b_n) = \lim \sup_{n \to \infty} g(b_n) \)，于是有\( f'(x_0) = \lim_{n \to \infty} g(b_n) \geq 0 \)。

如果\( f \)单调递减且\( f \)在\( x_0 \)可微，根据定理10.1.13的5，有 \( -f \)在\( x_0 \)可微，又\( -f \)是单调递增的，根据前面的证明，可得 \( (-f)'(x_0) = -f'(x_0) \geq 0 \)，进而可得\( f'(x_0) \leq 0 \)。

证毕。

练习10.3.2

题目：

Give an example of a function \( f: (−1, 1) \to \mathbf{R} \) which is continuous and monotone increasing, but which is not differentiable at \( 0 \). Explain why this does not contradict Proposition 10.3.1.

例子：

定义函数\( f: (-1, 1) \to \mathbf{R}, \forall x \in (-1, 1) \)，如果\( x \in (-1, 0) \)，则令\( f(x) := \dfrac{1}{2}x \)，如果\( x \in [0, 1) \)，则令\( f(x) := x \)，易证\( f \)连续且单调递增，但\( \lim_{x \to 0; x \in (-1, 0)} \dfrac{f(x) - f(0)}{x - 0} = \lim_{x \to 0; x \in (-1, 0)} \dfrac{\dfrac{1}{2}x - 0}{x - 0} = \dfrac{1}{2}, \lim_{x \to 0; x \in (0, 1)} \dfrac{f(x) - f(0)}{x - 0} = \lim_{x \to 0; x \in (0, 1)} \dfrac{x - 0}{x - 0} = 1 \neq \lim_{x \to 0; x \in (-1, 0)} \dfrac{f(x) - f(0)}{x - 0} \)，于是有\( \lim_{x \to 0; x \in (-1, 1) - \{ 0 \}} \dfrac{f(x) - f(0)}{x - 0} \)不存在，即\( f \)在\( 0 \)处不可微。

解释上面例子为什么没有和定理10.3.1矛盾：

使用定理10.3.1的前提条件之一是\( f \)在\( x_0 \)处可微（上面例子中\( x_0 = 0 \)），而上面例子不满足该前提条件，无法使用定理10.2.6，故不矛盾。

练习10.3.3

题目：

Give an example of a function \( f: \mathbf{R} \to \mathbf{R} \) which is strictly monotone increasing and differentiable, but whose derivative at \( 0 \) is zero. Explain why this does not contradict Proposition 10.3.1 or Proposition 10.3.3. (Hint: look at Exercise 10.2.3.)

例子：

该练习的例子可以直接取练习10.2.3的例子，改下定义域就行，定义函数\( f: \mathbf{R} \to \mathbf{R}, \forall x \in \mathbf{R} \)，令\( f(x) := x^3 \)，易证\( f \)严格单调递增，根据练习10.1.5，有\( f \)在\( \mathbf{R} \)上可微且 \( f'(0) = 3 \times 0^2 = 0 \)。

解释上面例子为什么没有和定理10.3.1矛盾：

对\( f \)使用定理10.3.1只能得到结论\( f'(0) \geq 0 \)，这包含\( f'(0) = 0 \)的可能性。

解释上面例子为什么没有和定理10.3.3矛盾：

给定\( a < b \in \mathbf{R} \)，则\( f|_{[a, b]} \)也严格单调递增且\( f|_{[a, b]} \)在\( [a, b] \)上可微（后者通过练习10.1.1得到），同时还有\( (f|_{[a, b]})'(0) = f'(0) = 0 \)，但我们只知道\( f|_{[a, b]} \)在\( 0 \)处的导数为\( 0 \)，这并不意味着\( f|_{[a, b]} \)在整个区间\( [a, b] \)上的导数均为\( 0 \)，故不满足使用定理10.3.3的第3种情况的前提条件（以及其他情况的前提条件也不满足），无法使用定理10.3.3，故不矛盾。

练习10.3.4

题目：

Prove Proposition 10.3.3. (Hint: you do not have integrals or the fundamental theorem of calculus yet, so these tools cannot be used. However, one can proceed via the mean-value theorem, Corollary 10.2.9.)

Proposition 10.3.3的内容：

Let \( a < b \), and let \( f: [a, b] \to \mathbf{R} \) be a differentiable function. If \( f'(x) > 0 \) for all \( x \in [a, b] \), then \( f \) is strictly monotone increasing. If \( f'(x) < 0 \) for all \( x \in [a, b] \), then \( f \) is strictly monotone decreasing. If \( f'(x) = 0 \) for all \( x \in [a, b] \), then \( f \) is a constant function.

证明：

如果\( \forall x \in [a, b], f'(x) > 0 \)，则\( \forall x < y \in [a, b] \)，根据\( f \)在\( [a, b] \)上可微、\( [x, y] \subseteq [a, b] \)以及练习10.1.1，可得\( f|_{[x, y]} \)在\( [x, y] \)上可微，进而根据定理10.1.10，可得\( f|_{[x, y]} \)在\( [x, y] \)上连续，根据推论10.2.9，可得\( \exists z \in [x, y], (f|_{[x, y]})'(z) = \dfrac{f(y) - f(x)}{y - x} \)，又\( (f|_{[x, y]})'(z) = f'(z) > 0 \)，可得\( \dfrac{f(y) - f(x)}{y - x} > 0 \)，进而可得\( f(y) - f(x) > y - x > 0 \)。综上，有\( f \)严格单调递增。

如果\( \forall x \in [a, b], f'(x) < 0 \)，则\( \forall x \in [a, b], (-f)'(x) = -f'(x) > 0 \)，根据前面的证明，可得\( -f \)严格单调递增，进而可得\( f \)严格单调递减。

如果\( \forall x \in [a, b], f'(x) = 0 \)，（这种情况跟\( f'(x) > 0 \)的证明方法类似，为了证明简短点可以写易证），则\( \forall x \neq y \in [a, b] \)，有\( x < y \)或\( y < x \)，不妨设\( x < y \)，此时根据\( f \)在\( [a, b] \)上可微、\( [x, y] \subseteq [a, b] \)以及练习10.1.1，可得\( f|_{[x, y]} \)在\( [x, y] \)上可微，进而根据定理10.1.10，可得\( f|_{[x, y]} \)在\( [x, y] \)上连续，根据推论10.2.9，可得\( \exists z \in [x, y], (f|_{[x, y]})'(z) = \dfrac{f(y) - f(x)}{y - x} \)，又\( (f|_{[x, y]})'(z) = f'(z) = 0 \)，可得\( \dfrac{f(y) - f(x)}{y - x} = 0 \)，进而可得\( f(y) - f(x) = 0 \)，即\( f(y) = f(x) \)。综上，有定义域内所有点的函数值都相等，因此\( f \)为常数函数。

证毕。

练习10.3.5

题目：

Give an example of a subset \( X \subseteq R \) and a function \( f: X \to \mathbf{R} \) which is differentiable on \( X \), is such that \( f'(x) > 0 \) for all \( x \in X \), but \( f \) is not strictly monotone increasing. (Hint: the conditions here are subtly different from those in Proposition 10.3.3. What is the difference, and how can one exploit that difference to obtain the example?)

例子：

这里的关键在于\( X \)可以不是闭区间，从而无法使用定理10.3.3。

令\( X := [0, 1] \cup [2, 3] \)，定义函数\( f: X \to \mathbf{R}, \forall x \in X \)，如果\( x \in [0, 1] \)，则令\( f(x) := x \)，如果\( x \in [2, 3] \)，则令\( f(x) := x - 2 \)。易证\( \forall x \in [0, 1], f'(x) = 1 > 0, \forall x \in [2, 3], f'(x) = 1 > 0 \)，即\( f \)在\( X \)上可微且\( \forall x \in X, f'(x) > 0 \)，但\( 2 > 1, f(2) = 0 < 1 = f(1) \)，因此\( f \)非严格单调递增。

章节10.4

练习10.4.1

题目：

Let \( n \geq 1 \) be a natural number, and let \( g : (0, +\infty) \to (0, +\infty) \) be the function \( g(x) := x^{1/n} \).

1. Show that \( g \) is continuous on \( (0, +\infty) \). (Hint: use Proposition 9.4.11.)
2. Show that \( g \) is differentiable on \( (0, +\infty) \), and that \( g'(x) = \dfrac{1}{n} x^{(1/n) - 1} \) for all \( x \in (0, +\infty) \). (Hint: use the inverse function theorem and 1.)

证明1：

定义函数\( f: (0, +\infty) \to \mathbf{R}, \forall x \in (0, +\infty) \)，令\( f(x) := g(x) \)（也就是\( f, g \)只有值域不同，其他都相同），根据定理9.4.11，可得\( f \)连续，而\( f, g \)只有值域不同，故可得\( g \)也连续。

证毕。

证明2：

根据定理6.7.3（更具体的，引理5.6.9的4），可得\( g \)严格单调递增，因此有\( g \)单射，而\( \forall y \in (0, +\infty) \)，有\( y^n \in (0, +\infty), g(y^n) = (y^n)^{1/n} = y \)，因此有\( g \)满射，综上，有\( g \)双射，可得\( g \)有逆函数\( g^{-1}: (0, +\infty) \to (0, +\infty) \)，特别的，定义函数\( f: (0, +\infty) \to (0, +\infty), \forall y \in (0, +\infty) \)，令\( f(y) := y^n \)，则\( \forall x \in (0, +\infty), f(g(x)) = (x^{1/n})^n = x, g(f(y)) = (y^n)^{1/n} = y \)，可得\( f = g^{-1} \)。

根据练习10.1.1和练习10.1.5，易得\( f \)在\( (0, +\infty) \)上可微（练习10.1.5中的函数值域不同，但是不影响结论）且\( \forall y \in (0, +\infty), f'(y) = ny^{n - 1} \neq 0 \) (1) 。

综上，有\( \forall x \in (0, +\infty) \)，\( f \)在\( g(x) \)处可微，\( f'(g(x)) \neq 0 \)， \( g \)在\( f(g(x)) = x \)处连续，满足定理10.4.2的使用条件，可得\( g'(x) = \dfrac{1}{f'(g(x))} \)，进而根据(1)可得，\( g'(x) = \dfrac{1}{f'(g(x))} = \dfrac{1}{n(x^{1/n})^{n - 1}} = \dfrac{1}{n(x^{(n - 1)/n})} = \dfrac{1}{n} \dfrac{1}{x^{(n - 1)/n}} = \dfrac{1}{n} \dfrac{1}{x^{1 - (1/n)}} = \dfrac{1}{n} x^{-(1 - (1/n))} = \dfrac{1}{n} x^{(1/n) - 1} \)。

证毕。

练习10.4.2

题目：

Let \( q \) be a rational number, and let \( f: (0, +\infty) \to \mathbf{R} \) be the function \( f(x) := x^q \).

1. Show that \( f \) is differentiable on \( (0, +\infty) \) and that \( f'(x) = qx^{q - 1} \). (Hint: use Exercise 10.4.1 and the laws of differential calculus in Theorem 10.1.13 and Theorem 10.1.15.)
2. Show that \( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{x^q - 1}{x - 1} = q \) for every rational number \( q \). (Hint: use part 1 and Definition 10.1.1. An alternate route is to apply L’Hôpital’s rule from the next section.)

证明1：

因为\( q \)是有理数，因此\( \exists a \geq 1 \in \mathbf{N}, b \in \mathbf{Z} \)，使得\( q = \dfrac{b}{a} \)。

定义函数\( g: (0, +\infty) \to (0, +\infty) \)， \( \forall x \in (0, +\infty) \)，令\( g(x) := x^{1/a} \)，根据练习10.4.1的2，可得\( g \)在\( x \)处可微且\( g'(x) = \dfrac{1}{a} x^{(1/a) - 1} \)。

定义函数\( h: (0, +\infty) \to \mathbf{R} \)， \( \forall y \in (0, +\infty) \)，令\( h(y) := y^b \)，根据练习10.1.1、练习10.1.5以及练习10.1.6，可得\( h \)在\( y \)处可微且\( h'(y) = by^{b - 1} \)。

综上，有\( \forall x \in (0, +\infty), f(x) = x^q = (x^{1/a})^b = h((g(x))) = (h \circ g)(x) \)，于是有\( f = h \circ g \)。针对\( \forall x \in (0, +\infty) \)，有\( g \)在\( x \)处可微，\( h \)在\( g(x) \)处可微，根据定理10.1.15，可得\( f'(x) = (h \circ g)'(x) = h'(g(x))g'(x) = (b(x^{1/a})^{b - 1})(\dfrac{1}{a} x^{(1/a) - 1}) = qx^{(b - 1)/a}x^{(1/a) - 1} = qx^{((b - 1)/a) + (1/a) - 1} = qx^{q - 1} \)。

证毕。

证明2：

根据1，可得\( f'(1) = q1^{q - 1} = q \)，根据定义10.1.1，这意味着 \( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{x^q - 1^q}{x - 1} = \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{x^q - 1}{x - 1} = f'(1) = q \)。

证毕。

练习10.4.3

题目：

Let \( \alpha \) be a real number, and let \( f: (0, +\infty) \to \mathbf{R} \) be the function \( f(x) = x^{\alpha} \).

1. Show that \( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{f(x) - f(1)}{x - 1} = \alpha \). (Hint: use Exercise 10.4.2 and the comparison principle; you may need to consider right and left limits separately. Proposition 5.4.14 may also be helpful.)
2. Show that \( f \) is differentiable on \( (0, +\infty) \) and that \( f'(x) = \alpha x^{\alpha - 1} \). (Hint: use 1, exponent laws (Proposition 6.7.3), and Definition 10.1.1.)

1的思路：

由定理9.3.9可知，为了证明\( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{f(x) - f(1)}{x - 1} = \alpha \)，我们仅需要证明\( \forall \)序列\( (a_n)_{n = 0}^{\infty} \) 满足\( \forall n \in \mathbf{N}, a_n \in (0, +\infty) - \{ 1 \} \)且\( \lim_{n \to \infty} a_n = 1 \)，均有\( \lim_{n \to \infty} \dfrac{f(a_n) - f(1)}{a_n - 1} = \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)即可。

下面的证明参考了 Validity of Limits? Proof that \( \lim_{x \to 1} \dfrac{x^{\alpha} - 1}{x - 1} = \alpha \) when \( \alpha \in \mathbf{R} \)中SurfaceIntegral的回答。

证明1：

针对\( \forall \)序列\( (a_n)_{n = 0}^{\infty} \) 满足\( \forall n \in \mathbf{N}, a_n \in (0, +\infty) - \{ 1 \} \)且\( \lim_{n \to \infty} a_n = 1 \)：

\( \forall m \in \mathbf{N} \)，由定理5.4.14，可得\( \exists q_m \in \mathbf{Q}, \alpha - \dfrac{1}{m} < q_m < \alpha \)，于是令\( Q_m := \{ q \in \mathbf{Q} : \alpha - \dfrac{1}{m} < q < \alpha \} \)，可得\( Q_m \)非空，使用选择公理，可以得到序列\( (q_m)_{m = 0}^{\infty} \)满足 \( \forall m \in \mathbf{N}, q_m \in Q_m \)，于是有\( q_m \in \mathbf{Q}, \alpha - \dfrac{1}{m} < q_m < \alpha \) (1) ，易证\( \lim_{m \to \infty} q_m = \alpha \)。

\( \forall m \in \mathbf{N} \)，由定理5.4.14，可得\( \exists p_m \in \mathbf{Q}, \alpha < p_m < \alpha + \dfrac{1}{m} \)，于是令\( P_m := \{ p \in \mathbf{Q} : \alpha < p < \alpha + \dfrac{1}{m} \} \)，可得\( P_m \)非空，使用选择公理，可以得到序列\( (p_m)_{m = 0}^{\infty} \)满足 \( \forall m \in \mathbf{N}, p_m \in P_m \)，于是有\( p_m \in \mathbf{Q}, \alpha < p_m < \alpha + \dfrac{1}{m} \) (2) ，易证\( \lim_{m \to \infty} p_m = \alpha \)。

\( \forall n, m \in \mathbf{N} \)，由于\( a_n \in (0, +\infty) - \{ 1 \} \)，因此可以分成\( 0 < a_n < 1, a_n > 1 \)两种情况讨论：

1. 如果\( a_n > 1 \)，则根据定理6.7.3（更具体的，引理5.6.9的5）以及(1)和(2)，可得\( (a_n)^{q_m} < (a_n)^{\alpha} < (a_n)^{p_m} \)，进而可得\( (a_n)^{q_m} - 1 < (a_n)^{\alpha} - 1 < (a_n)^{p_m} - 1 \)，而\( a_n > 1 \)，因此\( a_n - 1 > 0 \)，于是可得\( \dfrac{(a_n)^{q_m} - 1}{a_n - 1} < \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} < \dfrac{(a_n)^{p_m} - 1}{a_n - 1} \)。
2. 如果\( 0 < a_n < 1 \)，则根据定理6.7.3（更具体的，引理5.6.9的5）以及(1)和(2)，可得\( (a_n)^{p_m} < (a_n)^{\alpha} < (a_n)^{q_m} \)，进而可得\( (a_n)^{p_m} - 1 < (a_n)^{\alpha} - 1 < (a_n)^{q_m} - 1 \)，而\( 0 < a_n < 1 \)，因此\( a_n - 1 < 0 \)，于是可得\( \dfrac{(a_n)^{q_m} - 1}{a_n - 1} < \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} < \dfrac{(a_n)^{p_m} - 1}{a_n - 1} \)，这和情况1一样。

综上，\( \forall n, m \in \mathbf{N} \)，不管什么情况下均有 \( \dfrac{(a_n)^{q_m} - 1}{a_n - 1} < \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} < \dfrac{(a_n)^{p_m} - 1}{a_n - 1} \)，此时根据引理6.4.13，可得\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{q_m} - 1}{a_n - 1} \leq \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \leq \lim \sup_{n \to \infty} \dfrac{(a_n)^{p_m} - 1}{a_n - 1} \) (3) 。

\( \forall m \in \mathbf{N}, q_m \in \mathbf{Q} \)，根据练习10.4.2，有\( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{x^{q_m} - 1}{x - 1} = q_m \)，同理可得，\( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{x^{p_m} - 1}{x - 1} = p_m \)，进而根据定理9.3.9，可得\( \lim_{n \to \infty} \dfrac{(a_n)^{q_m} - 1}{a_n - 1} = q_m \) 以及\( \lim_{n \to \infty} \dfrac{(a_n)^{p_m} - 1}{a_n - 1} = p_m \)，再根据定理6.4.12的6，有\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{q_m} - 1}{a_n - 1} = q_m \) 以及\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{p_m} - 1}{a_n - 1} = p_m \)，于是根据(3)，有\( q_m \leq \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \leq p_m \) (4) ，这意味着\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \)是实数（而不是\( +\infty \)或者\( -\infty \)）。

定义序列\( (c_m)_{n = 0}^{\infty}, \forall m \in \mathbf{N} \)，令\( c_m := \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \)，（前面证明了\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \)是实数，所以这是合法的），可得\( (c_m)_{n = 0}^{\infty} \)为常数序列，针对该常数序列，可得\( \lim_{m \to \infty} c_m = \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} \) (5) ，而由(4)，有\( \forall m \in \mathbf{N}, q_m \leq c_m \leq p_m \)，又\( \lim_{m \to \infty} q_m = \alpha = \lim_{m \to \infty} p_m \)，于是根据推论6.4.14，可得\( \lim_{m \to \infty} c_m = \alpha \)，结合(5)，可得\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)。

同理易证\( \lim \inf_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)，于是有\( \lim \sup_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \lim \inf_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)，根据定理6.4.12的6，可得\( \lim_{n \to \infty} \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)。

综上，\( \forall \)序列\( (a_n)_{n = 0}^{\infty} \) 满足\( \forall n \in \mathbf{N}, a_n \in (0, +\infty) - \{ 1 \} \)且\( \lim_{n \to \infty} a_n = 1 \)，均有\( \lim_{n \to \infty} \dfrac{f(a_n) - f(1)}{a_n - 1} = \dfrac{(a_n)^{\alpha} - 1}{a_n - 1} = \alpha \)，于是根据定理9.3.9，可得\( \lim_{x \to 1; x \in (0, +\infty) - \{ 1 \}} \dfrac{f(x) - f(1)}{x - 1} = \alpha \)。

证毕。

证明2：

由1可得，\( f \)在\( 1 \)处可微且\( f'(1) = \alpha \)。

\( \forall x_0 \in (0, +\infty) \)，定义函数\( g_{x_0}: (0, +\infty) \to (0, +\infty), \forall x \in (0, +\infty) \)，令\( g_{x_0}(x) := \dfrac{x}{x_0} \)，根据定理10.1.13的5，可得\( g_{x_0} \)在\( (0, +\infty) \)上可微且\( (g_{x_0})'(x) = \dfrac{1}{x_0} \)，特别的，有\( g_{x_0} \)在\( x_0 \)处可微且\( (g_{x_0})'(x_0) = \dfrac{1}{x_0} \)，又\( g_{x_0}(x_0) = 1 \)， \( f \)在\( 1 \)处可微，根据定理10.1.15，可得\( f \circ g_{x_0} \)在\( x_0 \)处可微且 \( (f \circ g_{x_0})'(x_0) = f'(g_{x_0}(x_0))(g_{x_0})'(x_0) = f'(1)(g_{x_0})'(x_0) = \dfrac{\alpha}{x_0} \)，进而可得\( \lim_{x \to x_0; x \in (0, +\infty) - \{ x_0 \}} \dfrac{(\dfrac{x}{x_0})^{\alpha} - (\dfrac{x_0}{x_0})^{\alpha}}{x - x_0} = \lim_{x \to x_0; x \in (0, +\infty) - \{ x_0 \}} \dfrac{\dfrac{x^{\alpha}}{(x_0)^{\alpha}} - 1}{x - x_0} = \dfrac{\alpha}{x_0} \)，再次根据定理10.1.13的5，可得 \( \lim_{x \to x_0; x \in (0, +\infty) - \{ x_0 \}} (x_0)^{\alpha} \dfrac{\dfrac{x^{\alpha}}{(x_0)^{\alpha}} - 1}{x - x_0} = \lim_{x \to x_0; x \in (0, +\infty) - \{ x_0 \}} \dfrac{x^{\alpha} - (x_0)^{\alpha}}{x - x_0} = (x_0)^{\alpha} \dfrac{\alpha}{x_0} = \alpha (x_0)^{\alpha - 1} \)，即\( f \)在\( x_0 \)处可微且\( f'(x_0) = \alpha (x_0)^{\alpha - 1} \)。

综上，有\( f \)在\( (0, +\infty) \)上可微且\( \forall x \in (0, +\infty), f'(x) = \alpha x^{\alpha - 1} \)。

证毕。

章节10.5

练习10.5.1

题目：

Prove Proposition 10.5.1. (Hint: to show that \( g(x) \neq 0 \) near \( x_0 \), you may wish to use Newton’s approximation (Proposition 10.1.7). For the rest of the proposition, use limit laws, Proposition 9.3.14.)

Proposition 10.5.1的内容：

(L’Hôpital’s rule I). Let \( X \) be a subset of \( \mathbf{R} \), let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions, and let \( x_0 \in X \) be a limit point of \( X \). Suppose that \( f(x_0) = g(x_0) = 0 \), that \( f \) and \( g \) are both differentiable at \( x_0 \), but \( g'(x_0) \neq 0 \). Then there exists a \( \delta > 0 \) such that \( g(x) \neq 0 \) for all \( x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \} \), and \( \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{f(x)}{g(x)} = \dfrac{f'(x_0)}{g'(x_0)} \).

证明：

由\( g \)在\( x_0 \)处可微以及定理10.1.7，可得 \( \exists \delta > 0, \forall x \in X, |x - x_0| \leq \delta, |g(x) - g(x_0) - g'(x_0)(x - x_0)| \leq |\dfrac{g'(x_0)}{2}||x - x_0| \)，又\( g(x_0) = 0 \)，因此可得\( |g(x) - g'(x_0)(x - x_0)| \leq |\dfrac{g'(x_0)}{2}||x - x_0| \)。特别的，针对\( \forall x \in X - \{ x_0 \}, |x - x_0| < \delta \)，有\( x \in X, |x - x_0| \leq \delta \)，于是也有\( |g(x) - g'(x_0)(x - x_0)| \leq |\dfrac{g'(x_0)}{2}||x - x_0| \)，可得\( -|\dfrac{g'(x_0)}{2}||x - x_0| \leq g(x) - g'(x_0)(x - x_0) \leq |\dfrac{g'(x_0)}{2}||x - x_0| \)，进而可得\( g'(x_0)(x - x_0) - |\dfrac{g'(x_0)}{2}||x - x_0| \leq g(x) \leq g'(x_0)(x - x_0) + |\dfrac{g'(x_0)}{2}||x - x_0| \) (1) ，因为\( x \in X - \{ x_0 \} \)，因此\( x - x_0 \neq 0 \)，又\( g'(x_0) \neq 0 \)，因此\( g'(x_0)(x - x_0) \neq 0 \)，于是可以分\( g'(x_0)(x - x_0) > 0, g'(x_0)(x - x_0) < 0 \)两种情况讨论：

1. 如果\( g'(x_0)(x - x_0) > 0 \)，则\( |g'(x_0)||x - x_0| = |g'(x_0)(x - x_0)| = g'(x_0)(x - x_0) \)，于是根据(1)，可得\( (|g'(x_0)| - |\dfrac{g'(x_0)}{2}|)|x - x_0| \leq g(x) \leq (|g'(x_0)| + |\dfrac{g'(x_0)}{2}|)|x - x_0| \)，化简下就是\( |\dfrac{g'(x_0)}{2}||x - x_0| \leq g(x) \leq |\dfrac{3g'(x_0)}{2}||x - x_0| \)，不等式两边都\( > 0 \)，因此\( g(x) \neq 0 \)。
2. 如果\( g'(x_0)(x - x_0) < 0 \)，则\( -|g'(x_0)||x - x_0| = -|g'(x_0)(x - x_0)| = g'(x_0)(x - x_0) \)，于是根据(1)，可得\( (-|g'(x_0)| - |\dfrac{g'(x_0)}{2}|)|x - x_0| \leq g(x) \leq (-|g'(x_0)| + |\dfrac{g'(x_0)}{2}|)|x - x_0| \)，化简下就是\( -|\dfrac{3g'(x_0)}{2}||x - x_0| \leq g(x) \leq -|\dfrac{g'(x_0)}{2}||x - x_0| \)，不等式两边都\( < 0 \)，因此\( g(x) \neq 0 \)。

综上，\( \exists \delta > 0, \forall x \in X - \{ x_0 \}, |x - x_0| < \delta, g(x) \neq 0 \)，即\( \exists \delta > 0, \forall x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}, g(x) \neq 0 \)。

因为\( f, g \)在\( x_0 \)可微，因此有\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = f'(x_0) \) 以及\( \lim_{x \to x_0; x \in X - \{ x_0 \}} \dfrac{g(x) - g(x_0)}{x - x_0} = g'(x_0) \)，再根据定理9.3.8，可得\( \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{x - x_0} = f'(x_0) \) 以及\( \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{g(x) - g(x_0)}{x - x_0} = g'(x_0) \)，又\( \forall x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}, g(x) \neq 0 \)，于是根据定理9.3.14的除法法则以及\( f(x_0) = g(x_0) = 0 \)，可得\( \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{\dfrac{f(x) - f(x_0)}{x - x_0}}{\dfrac{g(x) - g(x_0)}{x - x_0}} = \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} (\dfrac{f(x) - f(x_0)}{x - x_0}) \times (\dfrac{x - x_0}{g(x) - g(x_0)}) = \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{f(x) - f(x_0)}{g(x) - g(x_0)} = \lim_{x \to x_0; x \in (X \cap (x_0 - \delta, x_0 + \delta)) - \{ x_0 \}} \dfrac{f(x)}{g(x)} = \dfrac{f'(x_0)}{g'(x_0)} \)。

证毕。

关于定理10.5.2的附注

Proposition 10.5.2原文的内容：

(L’Hôpital’s rule II). Let \( a < b \) be real numbers, let \( f: [a, b] \to \mathbf{R} \) and \( g: [a, b] \to \mathbf{R} \) be functions which are differentiable on \( [a, b] \). Suppose that \( f(a) = g(a) = 0 \), that \( g' \) is non-zero on \( [a, b] \) (i.e., \( g'(x) \neq 0 \) for all \( x \in [a, b] \)), and \( \lim_{x \to a; x \in (a,b]} \dfrac{f'(x)}{g'(x)} \) exists and equals \( L \). Then \( g(x) \neq 0 \) for all \( x \in (a, b] \), and \( \lim_{x \to a; x \in (a, b]} \dfrac{f(x)}{g(x)} \) exists and equals \( L \).

作者在博客中的勘误说明了，\( f, g \)在\( [a, b] \)上可微的条件可以弱化成 \( f, g \)在\( [a, b] \)上连续，在\( (a, b] \)上可微，对应的， \( \forall x \in [a, b], g'(x) \neq 0 \)的条件可以弱化成\( \forall x \in (a, b], g'(x) \neq 0 \)，修改后的定理如下：

Proposition 10.5.2条件弱化后的版本：

(L’Hôpital’s rule II). Let \( a < b \) be real numbers, let \( f: [a, b] \to \mathbf{R} \) and \( g: [a, b] \to \mathbf{R} \) be functions which are continuous on \( [a, b] \) and differentiable on \( (a, b] \). Suppose that \( f(a) = g(a) = 0 \), that \( g' \) is non-zero on \( (a, b] \) (i.e., \( g'(x) \neq 0 \) for all \( x \in (a, b] \)), and \( \lim_{x \to a; x \in (a,b]} \dfrac{f'(x)}{g'(x)} \) exists and equals \( L \). Then \( g(x) \neq 0 \) for all \( x \in (a, b] \), and \( \lim_{x \to a; x \in (a, b]} \dfrac{f(x)}{g(x)} \) exists and equals \( L \).

练习10.5.2

题目：

Explain why Example 1.2.12 does not contradict either of the propositions in this section.

Example 1.2.12的内容：

(L’Hôpital’s rule). We are all familiar with the beautifully simple L’Hôpital’s rule \( \lim_{x \to x_0} \dfrac{f(x)}{g(x)} = \lim_{x \to x_0} \dfrac{f'(x_0)}{g'(x_0)} \), but one can still get led to incorrect conclusions if one applies it incorrectly. For instance, applying it to \( f(x) := x, g(x) := 1 + x \), and \( x_0 := 0 \) we would obtain \( \lim_{x \to 0} \dfrac{x}{1 + x} = \lim_{x \to 0} \dfrac{1}{1} = 1 \), but this is the incorrect answer, since \( \lim_{x \to 0} \dfrac{x}{1 + x} = \dfrac{0}{1 + 0} = 0 \). Of course, all that is going on here is that L’Hôpital’s rule is only applicable when both \( f(x) \) and \( g(x) \) go to zero as \( x \to x_0 \), a condition which was violated in the above example. But even when \( f(x) \) and \( g(x) \) do go to zero as \( x \to x_0 \) there is still a possibility for an incorrect conclusion. For instance, consider the limit \( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} \). Both numerator and denominator go to zero as \( x \to 0 \), so it seems pretty safe to apply L’Hôpital’s rule, to obtain \( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} = \lim_{x \to 0} \dfrac{2x \sin(x^{-4}) - 4x^{-3} \cos(x^{-4})}{1} = \lim_{x \to 0} 2x \sin(x^{-4}) - \lim_{x \to 0} 4x^{-3} \cos(x^{-4}) \). The first limit converges to zero by the squeeze test (since the function \( 2x \sin(x^{−4} ) \) is bounded above by \( 2|x| \) and below by \( −2|x| \), both of which go to zero at \( 0 \)). But the second limit is divergent (because \( x^{−3} \) goes to infinity as \( x \to 0 \), and \( \cos(x^{-4}) \) does not go to zero). So the limit \( \lim_{x \to 0} \dfrac{2x \sin(x^{-4}) - 4x^{-3} \cos(x^{-4})}{1} \) diverges. One might then conclude using L’Hôpital’s rule that \( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} \) also diverges; however we can clearly rewrite this limit as \( \lim_{x \to 0} x \sin(x^{-4}) \), which goes to zero when \( x \to 0 \) by the squeeze test again. This does not show that L’Hôpital’s rule is untrustworthy (indeed, it is quite rigorous; see Section 10.5), but it still requires some care when applied.

注：

实际上例子1.2.12已经把各个例子不能用洛必达法则的原因讲的挺清楚了，只不过用的语言有点不严格。

解答：

针对例子\( f(x) := x, g(x) := 1 + x, x_0 := 0 \)，不能用洛必达法则的原因是 \( g(x_0) \neq 0 \)，不满足使用定理10.5.1、定理10.5.2的前提条件。

针对例子\( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} \)，文中以错误的方式使用了洛必达法则， \( \lim_{x \to 0} \dfrac{2x \sin(x^{-4}) - 4x^{-3} \cos(x^{-4})}{1} \)不存在并不能说明 \( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} \)不存在，洛必达法则只有在 \( \lim_{x \to 0} \dfrac{2x \sin(x^{-4}) - 4x^{-3} \cos(x^{-4})}{1} \)存在的情况下，才有可能使用洛必达法则得到关于\( \lim_{x \to 0} \dfrac{x^2 \sin(x^{-4})}{x} \)的结论（前提是还满足使用洛必达法则的其他前提条件）。

参考文章

1. Validity of Limits? Proof that \( \lim_{x \to 1} \dfrac{x^{\alpha} - 1}{x - 1} = \alpha \) when \( \alpha \in \mathbf{R} \)中SurfaceIntegral的回答