中值定理（基础阶段理论部分）

发表: 4/22/2026 更新: 5/22/2026 字数: 0 字 时长: 0 分钟

罗尔中值定理

Th1. (Rolle)

If

① $f(x)\in c[a,b]；$
② $f(x)\mathrm{在}(a,b)\mathrm{内}\mathrm{可}\mathrm{导}$
③ $f(a)=f(b)$

then$\mathrm{\exists }\zeta \in (a,b)，\mathrm{使}{f}^{\prime }(\zeta )=0$

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证:

$f(x)\in c[a,b]\Rightarrow \mathrm{\exists }m，M.$
① $m=M$
$f(x)\equiv C_0，\mathrm{则}\mathrm{\forall }\zeta \in (a,b)，\mathrm{有}{f}^{\prime }(\zeta )=0；$
② $m<M$
$\because f(a)=f(b).$
$\therefore m，M\mathrm{至}\mathrm{少}\mathrm{一}\mathrm{个}\mathrm{在}(a,b)\mathrm{内}\mathrm{取}\mathrm{到}。$
$\mathrm{设}\mathrm{\exists }\zeta \in (a,b).$
$\mathrm{使}f(\zeta )=M\Rightarrow \zeta \mathrm{为}\mathrm{极}\mathrm{大}\mathrm{点}\Rightarrow f^{\prime }(\zeta )=0\mathrm{或}\mathrm{不}\mathrm{存}\mathrm{在}$
$\because f(x)\mathrm{可}\mathrm{导}$
$\therefore f^{\prime }(\zeta )=0$

$\mathrm{例}1.f(x)\in c[0,2],(0,2)\mathrm{内}\mathrm{可}\mathrm{导}.3f(0)=f(1)+2f(2).$$\mathrm{证}:\mathrm{\exists }\zeta \in (0,2),\mathrm{使}{f}^{\prime }(\zeta )=0.$

证：
$1^0.f(x)\in c[1,2]\Rightarrow f(x)\mathrm{在}[1,2]\mathrm{上}\mathrm{有}m,M.$
$3m\le f(1)+2f(2)\le 3M\Rightarrow m\le \frac{f(1)+2f(2)}{3}\le M.$
$\mathrm{\exists }c\in [1,2],\mathrm{使}f(c)=\frac{f(1)+2f(2)}{3}\Rightarrow f(1)+2f(2)=3f(c)$
$2^0.\because f(0)=f(c)$（罗尔中值定理）
$\therefore \mathrm{\exists }\zeta \in (0,c)\subset (0,2),\mathrm{使}{f}^{\prime }(\zeta )=0$

拉格朗日中值定理（狭义上的微分中值定理）

Th2. (Lagrange)⭐

If

① $f(x)\in c[a,b]；$
② $f(x)\mathrm{在}(a,b)\mathrm{内}\mathrm{可}\mathrm{导}$

then$\mathrm{\exists }\zeta \in (a,b),\mathrm{使}{f}^{\prime }(\zeta )=\frac{f(b)-f(a)}{b-a}$

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分析:

$L_{\mathrm{曲}}:y=f(x)$
$L_{AB}:y-f(a)=\frac{f(b)-f(a)}{b-a}(x-a)$
$\mathrm{即}{L}_{AB}:y=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$

证:

$\mathrm{令}\phi (x)=\mathrm{曲}-\mathrm{直}=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$
$\phi (x)\in c[a,b].\mathrm{在}(a,b)\mathrm{内}\mathrm{可}\mathrm{导}$（函数的加减乘不会改变其连续性）
$\mathrm{又}\phi (a)=\phi (b)=0$
$\therefore \mathrm{\exists }\zeta \in (a,b),\mathrm{使}{\phi (\zeta )}^{\prime }=0$
$\mathrm{而}{\phi }^{\prime }(x)=f^{\prime }(x)-\frac{f(b)-f(a)}{b-a}$
$\therefore f^{\prime }(\zeta )=\frac{f(b)-f(a)}{b-a}$

Notes:

① $\mathrm{如}\mathrm{果}f(a)=f(b)，\mathrm{则}L\Rightarrow R;(Rolle\mathrm{定}\mathrm{理}\subset Lagrange\mathrm{定}\mathrm{理})$
② $f^{\prime }(\zeta )=\frac{f(b)-f(a)}{b-a}\iff f(b)-f(a)=f^{\prime }(\zeta )(b-a)\iff f(b)-f(a)=f^{\prime }[a+(b-a)\theta ](b-a)(0<\theta <1)$
③ $f(x)\mathrm{可}\mathrm{导}.\mathrm{则}f(x)-f(a)=f^{\prime }(\zeta )(x-a)=f^{\prime }(a+(b-a)\theta )(x-a)(0<\theta <1)$

$\mathrm{例}2.\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=e.\mathrm{求}\underset{x\to \mathrm{\infty }}{lim}[f(x+1)-f(x-1)].$

解：
$1^0.f(x+1)-f(x-1)=2f^{\prime }(\zeta )(x-1<\zeta <x+1).$(Lagrange)
$2^0.\mathrm{原}\mathrm{式}=2\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(\zeta )=2\underset{\zeta \to \mathrm{\infty }}{lim}{f}^{\prime }(\zeta )=2e.$

柯西中值定理

Th3. (Cauchy)

If

① $f(x),g(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{连}\mathrm{续};$
② $f(x),g(x)\mathrm{在}(a,b)\mathrm{内}\mathrm{可}\mathrm{导};$
③ $g^{\prime }(x)\ne 0(a<x<b);$

then$\mathrm{\exists }\zeta \in (a,b)，\mathrm{使}\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime }(\zeta )}{g^{\prime }(\zeta )}$

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Notes:

① $Ifg(x)=x,\mathrm{则}C\Rightarrow L;(Lagrange\mathrm{定}\mathrm{理}\subset Cauchy\mathrm{定}\mathrm{理})$
② $g^{\prime }(x)\ne 0(a<x<b)\Rightarrow \{\begin{array}{c}{g}^{\prime }(\zeta )\ne 0;\\ g(b)-g(a)\ne 0;\end{array}$
③
$L:\phi (x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$
$C:\phi (x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]$

证:

$\mathrm{令}\phi (x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]$
$\phi (x)\in c[a,b],(a,b)\mathrm{内}\mathrm{可}\mathrm{导}$
$\because \phi (a)=\phi (b)=0$
$\therefore \mathrm{\exists }\zeta \in (a,b),\mathrm{使}{\phi }^{\prime }(\zeta )=0$(罗尔中值定理)
$\mathrm{而}{\phi }^{\prime }(x)=f^{\prime }(x)-\frac{f(b)-f(a)}{g(b)-g(a)}{g}^{\prime }(x)$
$\therefore f^{\prime }(\zeta )=\frac{f(b)-f(a)}{g(b)-g(a)}{g}^{\prime }(\zeta )$
$\because g^{\prime }(\zeta )\ne 0$
$\therefore \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime }(\zeta )}{g^{\prime }(\zeta )}$

泰勒展开

Th4. (Taylor)

If

$f(x)\mathrm{在}x=x_0\mathrm{领}\mathrm{域}\mathrm{内}n+1\mathrm{阶}\mathrm{可}\mathrm{导}$

then$f(x)=P_n(x)+R_n(x)$

$P_n(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{″}(x_0)}{2!}(x-x_0{)}^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$
$R_n(x)=\{\begin{array}{c}\frac{f^{(n+1)}(\zeta )}{(n+1)!}(x-x_0{)}^{(n+1)}——Lagrange\mathrm{型}(\zeta \mathrm{介}\mathrm{于}{x}_0\mathrm{与}x\mathrm{之}\mathrm{间})\\ o((x-x_0{)}^n)——\mathrm{皮}\mathrm{亚}\mathrm{诺}\mathrm{型}\end{array}$
$If{x}_0=0:$
$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\cdots +\frac{f^{(n)}(0)}{n!}{x}^n+R_n(x)$(麦克劳林公式)
$R_n(x)=\{\begin{array}{c}\frac{f^{(n+1)}(\zeta )}{(n+1)!}(x{)}^{(n+1)}——Lagrange\mathrm{型}(\zeta \mathrm{介}\mathrm{于}0\mathrm{与}x\mathrm{之}\mathrm{间})\\ o((x{)}^n)——\mathrm{皮}\mathrm{亚}\mathrm{诺}\mathrm{型}\end{array}$

记:

① $e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+o(x^n);$
② $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots +\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}+o(x^{2n+1});$
③ $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots +\frac{(-1{)}^n}{(2n)!}{x}^{2n}+o(x^{2n});$
④ $\frac{1}{1-x}=1+x+x^2+\cdots +x^n+o(x^n);$
⑤ $\frac{1}{1+x}=1-x+x^2+\cdots +(-1{)}^n{x}^n+o(x^n);$
⑥ $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots +\frac{(-1{)}^{(n-1)}}{n}{x}^n+o(x^n);$
⑦ $(1+x{)}^a=1+ax+\frac{a(a-1)}{2!}{x}^2+\cdots +\frac{a(a-1)\cdots (a-n+1)}{n!}{x}^n+o(x^n).$

洛必达法则

应用:洛必达法则(0/0、∞/∞)

Case1 0/0
If

① $f(x),g(x)\mathrm{在}x=x_0\mathrm{去}\mathrm{心}\mathrm{领}\mathrm{域}\mathrm{内}\mathrm{可}\mathrm{导}\mathrm{且}{g}^{\prime }(x)\ne 0;$
② $\underset{x\to x_0}{lim}f(x)=0,\underset{x\to x_0}{lim}g(x)=0;$
③ $\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=A$

then$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=A$

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Notes:

$\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=A$ 是 $\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=A$ 的充分非必要条件$(P\Rightarrow Q,Q\nRightarrow P)$.

若 $\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ 不存在，$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}$ 不一定不存在$(\mathrm{\neg }P\nRightarrow \mathrm{\neg }Q)$.

仅表明洛必达法则不适用.

Case2 ∞/∞
If

① $f(x),g(x)\mathrm{在}x=x_0\mathrm{去}\mathrm{心}\mathrm{领}\mathrm{域}\mathrm{内}\mathrm{可}\mathrm{导}\mathrm{且}{g}^{\prime }(x)\ne 0;$
② $\underset{x\to x_0}{lim}f(x)=\mathrm{\infty },\underset{x\to x_0}{lim}g(x)=\mathrm{\infty };$
③ $\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=A$

then$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=A$

简单证明

① 当 $\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=\frac{0}{0}$ 时，定义 $f(x_0)=0,g(x_0)=0$
② 则 $\frac{f(x)}{g(x)}=\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\frac{f^{\prime }(\zeta )}{g^{\prime }(\zeta )}$ (Cauchy)
③ 当 $x\to x_0$，因为 $\zeta \in (x,x_0)(\mathrm{或}(x_0,x))$，此时 $\zeta \to x_0$
④ 则 $\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=\underset{x\to x_0}{lim}\frac{f^{\prime }(\zeta )}{g^{\prime }(\zeta )}=\underset{\zeta \to x_0}{lim}\frac{f^{\prime }(\zeta )}{g^{\prime }(\zeta )}=\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$(变量替换)